For the AP: \(\dfrac{3}{2},\ \dfrac{1}{2},\ -\dfrac{1}{2},\ -\dfrac{3}{2},\ldots\), write the first term \(a\) and the common difference \(d\).
The first term is simply the first number in the list, and \(d\) is found from any two consecutive terms:
So \(a=\dfrac{3}{2}\) and \(d=-1\). (We could equally well have used the next pair, \(-\dfrac{1}{2}-\dfrac{1}{2}=-1\) — any two consecutive terms give the same \(d\) once we already know the list is an AP.)
Which of the following lists of numbers form an AP? If they form an AP, write the next two terms:
(i) \(4,\ 10,\ 16,\ 22,\ldots\) (ii) \(1,\ -1,\ -3,\ -5,\ldots\)
(iii) \(-2,\ 2,\ -2,\ 2,\ -2,\ldots\) (iv) \(1,\ 1,\ 1,\ 2,\ 2,\ 2,\ 3,\ 3,\ 3,\ldots\)
(i) \(a_{2}-a_{1}=10-4=6\), \(a_{3}-a_{2}=16-10=6\), \(a_{4}-a_{3}=22-16=6\) — the same value every time. AP, \(d=6\). Next two terms: \(22+6=28\) and \(28+6=34\).
(ii) \(a_{2}-a_{1}=-1-1=-2\), \(a_{3}-a_{2}=-3-(-1)=-2\), \(a_{4}-a_{3}=-5-(-3)=-2\) — equal. AP, \(d=-2\). Next two terms: \(-5+(-2)=-7\) and \(-7+(-2)=-9\).
(iii) \(a_{2}-a_{1}=2-(-2)=4\), but \(a_{3}-a_{2}=-2-2=-4\). Since \(a_{2}-a_{1}\ne a_{3}-a_{2}\), this list does NOT form an AP.
(iv) \(a_{2}-a_{1}=1-1=0\), \(a_{3}-a_{2}=1-1=0\), but \(a_{4}-a_{3}=2-1=1\). Here \(a_{2}-a_{1}=a_{3}-a_{2}\ne a_{4}-a_{3}\). This list does NOT form an AP.
In which of the following situations does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km, when the fare is ₹15 for the first km and ₹8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes ¼ of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.
(i) The fare for 1, 2, 3, … km is \(15,\ 15+8=23,\ 23+8=31,\ 31+8=39,\ldots\) — each term is obtained by adding the SAME fixed amount, ₹8, to the one before it. Yes, this is an AP, with \(a=15\) and \(d=8\).
(ii) If the cylinder starts with volume V, removing ¼ of what remains leaves ¾ of it each time, so the volumes are \(V,\ \dfrac{3}{4}V,\ \left(\dfrac{3}{4}\right)^{2}V,\ldots\) Each term is obtained by MULTIPLYING the one before it by ¾, not by adding a fixed number. No, this is not an AP (it is a geometric progression).
(iii) The cost of digging each successive metre is \(150,\ 150+50=200,\ 200+50=250,\ 250+50=300,\ldots\) — a fixed ₹50 is added each time. Yes, this is an AP, with \(a=150\) and \(d=50\).
(iv) At 8% compound interest, the amount at the end of each year is the previous amount MULTIPLIED by 1.08: \(10000,\ 10000(1.08),\ 10000(1.08)^{2},\ldots\) There is no fixed amount being added each year (the actual rupee increase gets bigger every year). No, this is not an AP (it is a geometric progression).
Write first four terms of the AP, when the first term \(a\) and the common difference \(d\) are given as follows:
(i) \(a=10,\ d=10\) (ii) \(a=-2,\ d=0\) (iii) \(a=4,\ d=-3\)
(iv) \(a=-1,\ d=\dfrac{1}{2}\) (v) \(a=-1.25,\ d=-0.25\)
Each term after the first is obtained by adding \(d\) to the term before it, so the four terms are \(a,\ a+d,\ a+2d,\ a+3d\).
(i) \(a=10,\ d=10\): terms are \(10,\ 20,\ 30,\ 40\).
(ii) \(a=-2,\ d=0\): terms are \(-2,\ -2,\ -2,\ -2\) (a constant list is still an AP, with \(d=0\)).
(iii) \(a=4,\ d=-3\): terms are \(4,\ 1,\ -2,\ -5\).
(iv) \(a=-1,\ d=\dfrac{1}{2}\): terms are \(-1,\ -\dfrac{1}{2},\ 0,\ \dfrac{1}{2}\).
(v) \(a=-1.25,\ d=-0.25\): terms are \(-1.25,\ -1.50,\ -1.75,\ -2.00\).
For the following APs, write the first term and the common difference:
(i) \(3,\ 1,\ -1,\ -3,\ldots\) (ii) \(-5,\ -1,\ 3,\ 7,\ldots\)
(iii) \(\dfrac{1}{3},\ \dfrac{5}{3},\ \dfrac{9}{3},\ \dfrac{13}{3},\ldots\) (iv) \(0.6,\ 1.7,\ 2.8,\ 3.9,\ldots\)
The first term \(a\) is simply the first number in the list; the common difference \(d\) is any later term minus the one right before it.
(i) \(a=3,\ d=1-3=-2\).
(ii) \(a=-5,\ d=-1-(-5)=4\).
(iii) \(a=\dfrac{1}{3},\ d=\dfrac{5}{3}-\dfrac{1}{3}=\dfrac{4}{3}\).
(iv) \(a=0.6,\ d=1.7-0.6=1.1\).
Which of the following are APs? If they form an AP, find the common difference \(d\) and write three more terms:
(i) \(2,\ 4,\ 8,\ 16,\ldots\) (ii) \(2,\ \dfrac{5}{2},\ 3,\ \dfrac{7}{2},\ldots\) (iii) \(-1.2,\ -3.2,\ -5.2,\ -7.2,\ldots\)
(iv) \(-10,\ -6,\ -2,\ 2,\ldots\) (v) \(3,\ 3+\sqrt{2},\ 3+2\sqrt{2},\ 3+3\sqrt{2},\ldots\) (vi) \(0.2,\ 0.22,\ 0.222,\ 0.2222,\ldots\)
(vii) \(0,\ -4,\ -8,\ -12,\ldots\) (viii) \(-\dfrac{1}{2},\ -\dfrac{1}{2},\ -\dfrac{1}{2},\ -\dfrac{1}{2},\ldots\) (ix) \(1,\ 3,\ 9,\ 27,\ldots\)
(x) \(a,\ 2a,\ 3a,\ 4a,\ldots\) (xi) \(a,\ a^{2},\ a^{3},\ a^{4},\ldots\) (xii) \(\sqrt{2},\ \sqrt{8},\ \sqrt{18},\ \sqrt{32},\ldots\)
(xiii) \(\sqrt{3},\ \sqrt{6},\ \sqrt{9},\ \sqrt{12},\ldots\) (xiv) \(1^{2},\ 3^{2},\ 5^{2},\ 7^{2},\ldots\) (xv) \(1^{2},\ 5^{2},\ 7^{2},\ 73,\ldots\)
For each list, check whether the difference between EVERY pair of consecutive terms is the same. If yes, it is an AP and that common value is \(d\); if the differences disagree even once, it is not an AP.
(i) Differences \(4-2=2,\ 8-4=4\) — not equal. Not an AP (this list doubles each time, it is a GP).
(ii) Differences \(\dfrac{5}{2}-2=\dfrac{1}{2},\ 3-\dfrac{5}{2}=\dfrac{1}{2},\ \dfrac{7}{2}-3=\dfrac{1}{2}\) — all equal. AP, \(d=\dfrac{1}{2}\). Next three terms: \(4,\ \dfrac{9}{2},\ 5\).
(iii) Differences are all \(-3.2-(-1.2)=-2\). AP, \(d=-2\). Next three terms: \(-9.2,\ -11.2,\ -13.2\).
(iv) Differences are all \(4\). AP, \(d=4\). Next three terms: \(6,\ 10,\ 14\).
(v) Each term adds \(\sqrt{2}\) to the one before it. AP, \(d=\sqrt{2}\). Next three terms: \(3+4\sqrt{2},\ 3+5\sqrt{2},\ 3+6\sqrt{2}\).
(vi) Differences \(0.22-0.2=0.02,\ 0.222-0.22=0.002\) — not equal (they keep shrinking). Not an AP.
(vii) Differences are all \(-4\). AP, \(d=-4\). Next three terms: \(-16,\ -20,\ -24\).
(viii) Every term is the same number, so every difference is \(0\). AP, \(d=0\) (a constant list is a valid, if uneventful, AP). Next three terms: \(-\dfrac{1}{2},\ -\dfrac{1}{2},\ -\dfrac{1}{2}\).
(ix) Differences \(3-1=2,\ 9-3=6\) — not equal. Not an AP (each term is 3 times the one before it, a GP).
(x) Differences are all \(2a-a=a\). AP, \(d=a\) (true for any fixed value of \(a\), including \(a=0\), where every term is 0). Next three terms: \(5a,\ 6a,\ 7a\).
(xi) Differences: \(a^{2}-a=a(a-1)\) and \(a^{3}-a^{2}=a^{2}(a-1)\). Their difference is \(a^{2}(a-1)-a(a-1)=a(a-1)^{2}\), which is zero only in the special cases \(a=0\) or \(a=1\) — for every other value of \(a\) the two differences are unequal. Not an AP (for a general value of \(a\)).
(xii) Simplify each surd first: \(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\), \(\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\), \(\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}\). So the list is really \(\sqrt{2},\ 2\sqrt{2},\ 3\sqrt{2},\ 4\sqrt{2},\ldots\), with every difference equal to \(\sqrt{2}\). AP, \(d=\sqrt{2}\). Next three terms: \(5\sqrt{2}=\sqrt{50},\ 6\sqrt{2}=\sqrt{72},\ 7\sqrt{2}=\sqrt{98}\).
(xiii) Simplify: \(\sqrt{9}=3\), and \(\sqrt{12}=2\sqrt{3}\); \(\sqrt{6}\) does not simplify. Differences: \(\sqrt{6}-\sqrt{3}\approx1.449-1.732\)… numerically \(\sqrt{6}\approx2.449,\ \sqrt{3}\approx1.732\), so the first difference \(\approx0.717\); the second difference is \(3-\sqrt{6}\approx0.551\). These are not equal. Not an AP (unlike part (xii), the numbers under the root here do not factor out a common surd).
(xiv) \(1^{2}=1,\ 3^{2}=9,\ 5^{2}=25,\ 7^{2}=49\). Differences \(9-1=8,\ 25-9=16\) — not equal (squares of an AP are not themselves an AP). Not an AP.
(xv) \(1^{2}=1,\ 5^{2}=25,\ 7^{2}=49\), and then plainly \(73\) (not a square here — read the list as the four NUMBERS \(1,\ 25,\ 49,\ 73\)). Differences: \(25-1=24,\ 49-25=24,\ 73-49=24\) — all equal! AP, \(d=24\), even though the first three terms are written as squares — a reminder to always check the actual VALUES, not how they happen to be written. Next three terms: \(97,\ 121,\ 145\).
Find the 10th term of the AP: \(2,\ 7,\ 12,\ldots\)
Here \(a=2,\ d=7-2=5,\ n=10\). Using \(a_{n}=a+(n-1)d\):
So the 10th term of the given AP is 47.
Which term of the AP: \(21,\ 18,\ 15,\ldots\) is \(-81\)? Also, is any term 0? Give reason for your answer.
Here \(a=21,\ d=18-21=-3\), and \(a_{n}=-81\); we have to find \(n\). Using \(a_{n}=a+(n-1)d\):
So the 35th term of the given AP is \(-81\).
Next, is there any \(n\) for which \(a_{n}=0\)? If so:
Since \(n=8\) is a positive integer, yes, the 8th term is 0.
Determine the AP whose 3rd term is 5 and the 7th term is 9.
We have:
Subtracting the first equation from the second eliminates \(a\):
Substituting back into \(a+2d=5\): \(a+2=5\Rightarrow a=3\).
So \(a=3,\ d=1\), and the required AP is \(3,\ 4,\ 5,\ 6,\ 7,\ldots\).
Check whether 301 is a term of the list of numbers \(5,\ 11,\ 17,\ 23,\ldots\)
First confirm the list is an AP: \(a_{2}-a_{1}=6,\ a_{3}-a_{2}=6,\ a_{4}-a_{3}=6\) — equal every time, so it is an AP with \(a=5,\ d=6\).
Suppose 301 IS a term, say the \(n\)th term. Then:
But \(n\) must be a positive INTEGER (a term number can only be 1st, 2nd, 3rd, …), and \(\dfrac{151}{3}\) is not a whole number. So 301 is NOT a term of the given list.
How many two-digit numbers are divisible by 3?
The two-digit numbers divisible by 3 are \(12,\ 15,\ 18,\ldots,\ 99\) — this is an AP with \(a=12,\ d=3,\ a_{n}=99\). Using \(a_{n}=a+(n-1)d\):
So, there are 30 two-digit numbers divisible by 3.
Find the 11th term from the last term (towards the first term) of the AP: \(10,\ 7,\ 4,\ldots,\ -62\).
Here \(a=10,\ d=7-10=-3,\ l=-62\), where \(l=a+(n-1)d\) is the last term.
Method: find the total number of terms first.
So there are 25 terms in the given AP. The 11th term FROM THE LAST will be the \((25-11+1)=15\)th term from the start (not the 14th — counting the last term itself as the 1st-from-the-last, the 11th-from-the-last is 10 steps further towards the start):
So, the 11th term from the last term is \(-32\).
Alternative Solution: write the given AP in REVERSE order. Then the first term becomes \(a=-62\) and the common difference becomes \(d=3\) (going from the last term back to the first, each step ADDS 3). The question now becomes: find the 11th term of THIS reversed AP.
The 11th term of the reversed AP, which is now the required term, is \(-32\) — matching the first method exactly.
A sum of ₹1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact.
Simple interest \(=\dfrac{P\times R\times T}{100}\). With \(P=1000,\ R=8\):
So the interest (in ₹) at the end of the 1st, 2nd, 3rd, … years is \(80,\ 160,\ 240,\ldots\) The difference between consecutive terms is always 80. Yes, this is an AP, with \(a=80,\ d=80\).
To find the interest at the end of 30 years, find \(a_{30}\):
So, the interest at the end of 30 years will be ₹2400.
In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
The number of rose plants in the 1st, 2nd, 3rd, … rows is \(23,\ 21,\ 19,\ldots,\ 5\), which forms an AP (each row has 2 fewer plants than the one before). Let the number of rows be \(n\). Then:
Using \(a_{n}=a+(n-1)d\):
So, there are 10 rows in the flower bed.
Fill in the blanks in the following table, given that \(a\) is the first term, \(d\) the common difference and \(a_{n}\) the \(n\)th term of the AP:
| a | d | n | an | |
|---|---|---|---|---|
| (i) | 7 | 3 | 8 | … |
| (ii) | −18 | … | 10 | 0 |
| (iii) | … | −3 | 18 | −5 |
| (iv) | −18.9 | 2.5 | … | 3.6 |
| (v) | 3.5 | 0 | 105 | … |
Every row uses the same relation \(a_{n}=a+(n-1)d\), with whichever three of the four quantities are known.
(i) \(a_{8}=7+(8-1)(3)=7+21=\mathbf{28}\).
(ii) \(0=-18+(10-1)d\Rightarrow9d=18\Rightarrow d=\mathbf{2}\).
(iii) \(-5=a+(18-1)(-3)=a-51\Rightarrow a=\mathbf{46}\).
(iv) \(3.6=-18.9+(n-1)(2.5)\Rightarrow(n-1)=\dfrac{22.5}{2.5}=9\Rightarrow n=\mathbf{10}\).
(v) \(a_{105}=3.5+(105-1)(0)=\mathbf{3.5}\) (with \(d=0\) every term is the first term).
Choose the correct choice and justify:
(i) 30th term of the AP \(10,\ 7,\ 4,\ldots\) is: (A) 97 (B) 77 (C) \(-77\) (D) \(-87\)
(ii) 11th term of the AP \(-3,\ -\dfrac{1}{2},\ 2,\ldots\) is: (A) 28 (B) 22 (C) \(-38\) (D) \(-48\dfrac{1}{2}\)
(i) \(a=10,\ d=7-10=-3\).
The correct choice is (C) \(-77\).
(ii) \(a=-3,\ d=-\dfrac{1}{2}-(-3)=\dfrac{5}{2}\).
The correct choice is (B) 22.
In the following APs, find the missing terms in the boxes:
(i) \(2,\ \square,\ 26\) (ii) \(\square,\ 13,\ \square,\ 3\) (iii) \(5,\ \square,\ \square,\ 9\dfrac{1}{2}\)
(iv) \(-4,\ \square,\ \square,\ \square,\ \square,\ 6\) (v) \(\square,\ 38,\ \square,\ \square,\ \square,\ -22\)
(i) Three terms \(2,\ \square,\ 26\): \(d=\dfrac{26-2}{2}=12\). Missing term \(=2+12=\mathbf{14}\).
(ii) Four terms with the 2nd \(=13\) and the 4th \(=3\): \(2d=3-13=-10\Rightarrow d=-5\). 1st term \(=13-d=13-(-5)=\mathbf{18}\); 3rd term \(=13+d=13-5=\mathbf{8}\).
(iii) \(t_{1}=5,\ t_{4}=9\dfrac{1}{2}=\dfrac{19}{2}\): \(d=\dfrac{\frac{19}{2}-5}{3}=\dfrac{9/2}{3}=\dfrac{3}{2}\). \(t_{2}=5+\dfrac{3}{2}=\mathbf{6\dfrac{1}{2}}\); \(t_{3}=5+2\left(\dfrac{3}{2}\right)=\mathbf{8}\).
(iv) \(t_{1}=-4,\ t_{6}=6\): \(d=\dfrac{6-(-4)}{5}=\dfrac{10}{5}=2\). Missing terms: \(t_{2}=-2,\ t_{3}=0,\ t_{4}=2,\ t_{5}=4\).
(v) \(t_{2}=38,\ t_{6}=-22\): \(d=\dfrac{-22-38}{4}=\dfrac{-60}{4}=-15\). \(t_{1}=38-d=38-(-15)=\mathbf{53}\); \(t_{3}=38+d=\mathbf{23}\); \(t_{4}=23+d=\mathbf{8}\); \(t_{5}=8+d=\mathbf{-7}\) (check: \(t_{6}=-7+(-15)=-22\ \checkmark\)).
Which term of the AP: \(3,\ 8,\ 13,\ 18,\ldots\), is 78?
Here \(a=3,\ d=5\). Using \(a_{n}=a+(n-1)d=78\):
So, 78 is the 16th term of the given AP.
Find the number of terms in each of the following APs:
(i) \(7,\ 13,\ 19,\ldots,\ 205\) (ii) \(18,\ 15\dfrac{1}{2},\ 13,\ldots,\ -47\)
(i) \(a=7,\ d=6,\ a_{n}=205\). \((n-1)=\dfrac{205-7}{6}=\dfrac{198}{6}=33\Rightarrow n=\mathbf{34}\).
(ii) \(a=18,\ d=15\dfrac{1}{2}-18=-\dfrac{5}{2},\ a_{n}=-47\). \((n-1)=\dfrac{-47-18}{-5/2}=\dfrac{-65}{-5/2}=26\Rightarrow n=\mathbf{27}\).
Check whether \(-150\) is a term of the AP: \(11,\ 8,\ 5,\ 2,\ldots\)
Here \(a=11,\ d=-3\). If \(-150\) were the \(n\)th term:
Since \(n\) does not come out to be a positive integer, \(-150\) is NOT a term of the given AP.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
\(a_{16}-a_{11}=5d=73-38=35\Rightarrow d=7\). Since \(a_{11}=a+10d=38\), \(a=38-70=-32\).
So, the 31st term is 178.
An AP consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.
Here \(a_{3}=a+2d=12\) and \(a_{50}=a+49d=106\). Subtracting:
Then \(a=12-2(2)=8\).
So, the 29th term is 64.
If the 3rd and the 9th terms of an AP are 4 and \(-8\) respectively, which term of this AP is zero?
\(a_{3}=a+2d=4\) and \(a_{9}=a+8d=-8\). Subtracting:
Then \(a=4-2(-2)=8\). Setting \(a_{n}=0\):
So, the 5th term of this AP is zero.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
“17th term exceeds 10th term by 7” means \(a_{17}-a_{10}=7\). Since \(a_{17}-a_{10}=(a+16d)-(a+9d)=7d\):
So, the common difference is 1.
Which term of the AP: \(3,\ 15,\ 27,\ 39,\ldots\) will be 132 more than its 54th term?
Here \(d=12\). We need \(a_{n}-a_{54}=132\). Since \(a_{n}-a_{54}=(n-54)d\):
So, the 65th term will be 132 more than the 54th term.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Let the two APs have first terms \(a\) and \(b\), and the SAME common difference \(d\). Then the difference between their \(n\)th terms is:
Notice that the \((n-1)d\) terms cancel completely — so the difference between the \(n\)th terms of two APs sharing a common difference is ALWAYS just \(a-b\), the same for every \(n\), regardless of which term number \(n\) is.
Since the difference between the 100th terms is 100, we have \(a-b=100\). By the same reasoning, the difference between the 1000th terms is also 100.
How many three-digit numbers are divisible by 7?
The smallest three-digit multiple of 7 is \(105\ (=7\times15)\), and the largest is \(994\ (=7\times142)\). So the list is an AP: \(a=105,\ d=7,\ a_{n}=994\).
So, there are 128 three-digit numbers divisible by 7.
How many multiples of 4 lie between 10 and 250?
The multiples of 4 strictly between 10 and 250 are \(12,\ 16,\ 20,\ldots,\ 248\) (the next multiple after 248 would be 252, which exceeds 250). This is an AP: \(a=12,\ d=4,\ a_{n}=248\).
So, there are 60 multiples of 4 between 10 and 250.
For what value of \(n\), are the \(n\)th terms of two APs: \(63,\ 65,\ 67,\ldots\) and \(3,\ 10,\ 17,\ldots\) equal?
For the first AP, \(a=63,\ d=2\), so \(a_{n}=63+2(n-1)=61+2n\).
For the second AP, \(a=3,\ d=7\), so \(b_{n}=3+7(n-1)=7n-4\).
Setting the two \(n\)th terms equal:
So, the 13th terms of the two APs are equal (both equal \(61+26=87\)). \(n=13\).
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
“7th term exceeds 5th term by 12” means \(a_{7}-a_{5}=12\). Since \(a_{7}-a_{5}=2d\):
Since \(a_{3}=a+2d=16\):
So, the required AP has \(a=4,\ d=6\): \(4,\ 10,\ 16,\ 22,\ 28,\ldots\)
Find the 20th term from the last term of the AP: \(3,\ 8,\ 13,\ldots,\ 253\).
First find the total number of terms: \(a=3,\ d=5,\ a_{n}=253\).
The 20th term from the last is the \((51-20+1)=32\)nd term from the start:
So, the 20th term from the last term is 158.
Quick check (reverse-AP method, as in Example 8): reversed, \(a=253,\ d=-5\); \(a_{20}=253+19(-5)=253-95=158\) — the same answer.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
\(a_{4}+a_{8}=(a+3d)+(a+7d)=2a+10d=24\ \Rightarrow\ a+5d=12\).
\(a_{6}+a_{10}=(a+5d)+(a+9d)=2a+14d=44\ \Rightarrow\ a+7d=22\).
Subtracting the first from the second:
Then \(a=12-5(5)=-13\).
So, the first three terms are \(-13,\ -8,\ -3\).
Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?
The salary in successive years is an AP with \(a=5000,\ d=200\). We need \(a_{n}=7000\):
So the 11th year of work is when his income reached ₹7000. Taking 1995 as year 1, the 11th year is \(1995+10=\)2005.
Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the \(n\)th week, her weekly savings become ₹20.75, find \(n\).
Here \(a=5,\ d=1.75,\ a_{n}=20.75\):
So, \(n=10\) — her savings reach ₹20.75 in the 10th week.
Find the sum of the first 22 terms of the AP: \(8,\ 3,\ -2,\ldots\)
Here \(a=8,\ d=3-8=-5,\ n=22\). Using \(S_{n}=\dfrac{n}{2}[2a+(n-1)d]\):
So, the sum of the first 22 terms of the AP is \(-979\).
If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Here \(S_{14}=1050,\ n=14,\ a=10\). Using \(S_{n}=\dfrac{n}{2}[2a+(n-1)d]\):
Now find \(a_{20}\):
So, the 20th term is 200.
How many terms of the AP: \(24,\ 21,\ 18,\ldots\) must be taken so that their sum is 78?
Here \(a=24,\ d=21-24=-3,\ S_{n}=78\); we have to find \(n\). Using \(S_{n}=\dfrac{n}{2}[2a+(n-1)d]\):
Both values are admissible, since \(n\) must simply be a positive integer and both 4 and 13 qualify. So, the number of terms is either 4 or 13.
Why two answers? In this AP, \(a=24\) is positive but \(d=-3\) is negative, so the terms start positive and eventually become negative. The sum of the first 4 terms equals the sum of the first 13 terms (both equal 78) precisely because the terms from the 5th to the 13th add up to exactly \(0\) — some of those terms are positive and some negative, and they cancel out.
Find the sum of: (i) the first 1000 positive integers (ii) the first \(n\) positive integers.
(i) Let \(S=1+2+3+\ldots+1000\). This is an AP with \(a=1,\ l=1000,\ n=1000\). Using \(S_{n}=\dfrac{n}{2}(a+l)\):
So, the sum of the first 1000 positive integers is 500500.
(ii) Let \(S_{n}=1+2+3+\ldots+n\). Here \(a=1\) and the last term \(l=n\), so:
So, the sum of the first \(n\) positive integers is \(\dfrac{n(n+1)}{2}\) — a formula worth remembering, since it is used repeatedly (Example 15 and several exercise questions below reuse it directly).
Find the sum of the first 24 terms of the list of numbers whose \(n\)th term is given by \(a_{n}=3+2n\).
Since \(a_{n}=3+2n\), the first few terms are \(a_{1}=3+2=5,\ a_{2}=3+4=7,\ a_{3}=3+6=9,\ldots\), giving the list \(5,\ 7,\ 9,\ 11,\ldots\). Here \(7-5=9-7=11-9=2\), so it forms an AP with common difference \(d=2\).
To find \(S_{24}\), we have \(n=24,\ a=5,\ d=2\):
So, the sum of the first 24 terms of the given list of numbers is 672.
A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find: (i) the production in the 1st year (ii) the production in the 10th year (iii) the total production in the first 7 years.
(i) Since the production increases by a fixed number every year, the number of TV sets in year 1, 2, 3, … forms an AP. Let \(a_{n}\) denote the number of sets produced in year \(n\). We are given \(a_{3}=600\) and \(a_{7}=700\):
Subtracting, \(4d=100\Rightarrow d=25\); substituting back, \(a=600-2(25)=550\). So, production in the first year is 550 sets.
(ii) \(a_{10}=a+9d=550+9\times25=550+225=775\). So, production in the 10th year is 775 sets.
(iii) \(S_{7}=\dfrac{7}{2}\left[2(550)+(7-1)(25)\right]=\dfrac{7}{2}\left[1100+150\right]=\dfrac{7}{2}(1250)=4375\). So, the total production in the first 7 years is 4375 sets.
Find the sum of the following APs:
(i) \(2,\ 7,\ 12,\ldots\), to 10 terms.
(ii) \(-37,\ -33,\ -29,\ldots\), to 12 terms.
(iii) \(0.6,\ 1.7,\ 2.8,\ldots\), to 100 terms.
(iv) \(\dfrac{1}{15},\ \dfrac{1}{12},\ \dfrac{1}{10},\ldots\), to 11 terms.
In every part, use \(S_{n}=\dfrac{n}{2}\left[2a+(n-1)d\right]\).
(i) \(a=2,\ d=5,\ n=10\).
(ii) \(a=-37,\ d=4,\ n=12\).
(iii) \(a=0.6,\ d=1.1,\ n=100\).
(iv) \(a=\dfrac{1}{15},\ d=\dfrac{1}{12}-\dfrac{1}{15}=\dfrac{1}{60},\ n=11\) (check with the next pair too: \(\dfrac{1}{10}-\dfrac{1}{12}=\dfrac{1}{60}\ \checkmark\)).
So the four sums are 245, \(-180\), 5505, \(\dfrac{33}{20}\) respectively.
Find the sums given below:
(i) \(7+10\dfrac{1}{2}+14+\ldots+84\)
(ii) \(34+32+30+\ldots+10\)
(iii) \(-5+(-8)+(-11)+\ldots+(-230)\)
In each part, the number of terms \(n\) is not given directly, so first find it from \(n=\dfrac{l-a}{d}+1\), then use \(S_{n}=\dfrac{n}{2}(a+l)\).
(i) \(a=7,\ d=3\dfrac{1}{2},\ l=84\).
(ii) \(a=34,\ d=-2,\ l=10\).
(iii) \(a=-5,\ d=-3,\ l=-230\).
So the three sums are 1046.5, 286, \(-8930\) respectively.
In an AP:
(i) given \(a=5,\ d=3,\ a_{n}=50\), find \(n\) and \(S_{n}\).
(ii) given \(a=7,\ a_{13}=35\), find \(d\) and \(S_{13}\).
(iii) given \(a_{12}=37,\ d=3\), find \(a\) and \(S_{12}\).
(iv) given \(a_{3}=15,\ S_{10}=125\), find \(d\) and \(a_{10}\).
(v) given \(d=5,\ S_{9}=75\), find \(a\) and \(a_{9}\).
(vi) given \(a=2,\ d=8,\ S_{n}=90\), find \(n\) and \(a_{n}\).
(vii) given \(a=8,\ a_{n}=62,\ S_{n}=210\), find \(n\) and \(d\).
(viii) given \(a_{n}=4,\ d=2,\ S_{n}=-14\), find \(n\) and \(a\).
(ix) given \(a=3,\ n=8,\ S=192\), find \(d\).
(x) given \(l=28,\ S=144\), and there are a total of 9 terms. Find \(a\).
(i) \(a_{n}=5+(n-1)3=50\Rightarrow(n-1)=15\Rightarrow\mathbf{n=16}\). \(S_{16}=\dfrac{16}{2}(5+50)=8(55)=\mathbf{440}\).
(ii) \(a_{13}=a+12d=35\Rightarrow7+12d=35\Rightarrow\mathbf{d=\dfrac{28}{12}=\dfrac{7}{3}}\). \(S_{13}=\dfrac{13}{2}(7+35)=\dfrac{13}{2}(42)=\mathbf{273}\).
(iii) \(a_{12}=a+11d=37\Rightarrow a=37-33=\mathbf{4}\). \(S_{12}=\dfrac{12}{2}(a+a_{12})=6(4+37)=\mathbf{246}\).
(iv) From \(a_{3}=a+2d=15\), \(a=15-2d\). Substituting into \(S_{10}=\dfrac{10}{2}\left[2a+9d\right]=5(2a+9d)=125\), i.e. \(2a+9d=25\):
Then \(a=15-2(-1)=17\), and \(\mathbf{a_{10}=a+9d=17-9=8}\).
(v) \(S_{9}=\dfrac{9}{2}\left[2a+8d\right]=9(a+4d)=75\Rightarrow a+4d=\dfrac{25}{3}\). With \(d=5\): \(\mathbf{a=\dfrac{25}{3}-20=-\dfrac{35}{3}}\), and \(\mathbf{a_{9}=a+8d=-\dfrac{35}{3}+40=\dfrac{85}{3}}\).
(vi) \(S_{n}=\dfrac{n}{2}\left[4+8(n-1)\right]=n(4n-2)=90\Rightarrow4n^{2}-2n-90=0\Rightarrow2n^{2}-n-45=0\). By the quadratic formula, \(n=\dfrac{1\pm\sqrt{1+360}}{4}=\dfrac{1\pm19}{4}\); taking the positive root, \(\mathbf{n=5}\). Then \(\mathbf{a_{5}=2+4(8)=34}\).
(vii) \(S_{n}=\dfrac{n}{2}(a+a_{n})=\dfrac{n}{2}(70)=35n=210\Rightarrow\mathbf{n=6}\). Then \(a_{6}=8+5d=62\Rightarrow\mathbf{d=\dfrac{54}{5}=10.8}\).
(viii) From \(a_{n}=a+(n-1)(2)=4\), \(a=6-2n\). Substituting into \(S_{n}=\dfrac{n}{2}(a+a_{n})=\dfrac{n}{2}\left[(6-2n)+4\right]=\dfrac{n}{2}(10-2n)=n(5-n)=-14\):
Then \(\mathbf{a=6-2(7)=-8}\).
(ix) \(S_{8}=\dfrac{8}{2}\left[2(3)+7d\right]=4(6+7d)=192\Rightarrow6+7d=48\Rightarrow\mathbf{d=6}\).
(x) \(S_{9}=\dfrac{9}{2}(a+28)=144\Rightarrow a+28=32\Rightarrow\mathbf{a=4}\).
How many terms of the AP: \(9,\ 17,\ 25,\ldots\) must be taken to give a sum of 636?
\(a=9,\ d=8,\ S_{n}=636\).
By the quadratic formula, with discriminant \(5^{2}+4(4)(636)=25+10176=10201=101^{2}\):
(The negative root is rejected since \(n\) must be a positive integer.) So, 12 terms must be taken.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
\(a=5,\ l=45,\ S=400\). Using \(S_{n}=\dfrac{n}{2}(a+l)\):
Now find \(d\) from \(l=a+(n-1)d\):
So, there are 16 terms, with common difference \(\dfrac{8}{3}\).
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
\(a=17,\ l=350,\ d=9\). Using \(l=a+(n-1)d\):
Now the sum, using \(S_{n}=\dfrac{n}{2}(a+l)\):
So, there are 38 terms, with sum 6973.
Find the sum of first 22 terms of an AP in which \(d=7\) and 22nd term is 149.
\(a_{22}=a+21d=149\Rightarrow a=149-21(7)=149-147=2\).
So, the sum of the first 22 terms is 1661.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
\(d=a_{3}-a_{2}=18-14=4\); \(a=a_{2}-d=14-4=10\).
So, the sum of the first 51 terms is 5610.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first \(n\) terms.
\(S_{7}=\dfrac{7}{2}\left[2a+6d\right]=7(a+3d)=49\Rightarrow a+3d=7\).
\(S_{17}=\dfrac{17}{2}\left[2a+16d\right]=17(a+8d)=289\Rightarrow a+8d=17\).
Subtracting the first equation from the second:
Then \(a=7-3(2)=1\). Now:
So, the sum of the first \(n\) terms is \(S_{n}=n^{2}\) — a strikingly clean result, which follows because \(a=1,\ d=2\) makes this the AP of ODD numbers \(1,3,5,7,\ldots\), whose sum of the first \(n\) terms is always a perfect square.
Show that \(a_{1},\ a_{2},\ldots,\ a_{n},\ldots\) form an AP where \(a_{n}\) is defined as below:
(i) \(a_{n}=3+4n\) (ii) \(a_{n}=9-5n\)
Also find the sum of the first 15 terms in each case.
(i) \(a_{n}=3+4n\): \(a_{1}=7,\ a_{2}=11,\ a_{3}=15,\ldots\) In general, \(a_{n+1}-a_{n}=\left[3+4(n+1)\right]-\left[3+4n\right]=4\), the SAME value for every \(n\). So this is an AP, with \(a=7,\ d=4\).
(ii) \(a_{n}=9-5n\): \(a_{1}=4,\ a_{2}=-1,\ a_{3}=-6,\ldots\) In general, \(a_{n+1}-a_{n}=\left[9-5(n+1)\right]-\left[9-5n\right]=-5\), again the same for every \(n\). So this is an AP, with \(a=4,\ d=-5\).
If the sum of the first \(n\) terms of an AP is \(4n-n^{2}\), what is the first term (that is \(S_{1}\))? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the \(n\)th terms.
The first term is simply \(S_{1}\):
The sum of the first two terms is \(S_{2}\):
Since \(S_{2}=a_{1}+a_{2}\), the second term is:
So the common difference is \(d=a_{2}-a_{1}=1-3=-2\). Using the general rule \(a_{n}=S_{n}-S_{n-1}\) for \(n\ge2\):
Check: \(a_{1}=5-2=3\ \checkmark\), \(a_{2}=5-4=1\ \checkmark\). So:
So, \(a_{1}=3\), \(S_{2}=4\), \(a_{2}=1\), \(a_{3}=-1\), \(a_{10}=-15\), and in general \(a_{n}=5-2n\).
Find the sum of the first 40 positive integers divisible by 6.
The list is \(6,\ 12,\ 18,\ldots\), 40 terms, with \(a=6,\ d=6\).
So, the sum of the first 40 positive integers divisible by 6 is 4920.
Find the sum of the first 15 multiples of 8.
The list is \(8,\ 16,\ 24,\ldots\), 15 terms, with \(a=8,\ d=8\).
So, the sum of the first 15 multiples of 8 is 960.
Find the sum of the odd numbers between 0 and 50.
The odd numbers strictly between 0 and 50 are \(1,\ 3,\ 5,\ldots,\ 49\), with \(a=1,\ d=2,\ l=49\).
So, the sum of the odd numbers between 0 and 50 is 625.
A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹50 more than for the preceding day. How much money does the contractor have to pay as penalty, if he has delayed the work by 30 days?
The penalty per day forms an AP: \(a=200,\ d=50,\ n=30\).
So, the contractor has to pay a total penalty of ₹27,750.
A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.
Let the largest (first) prize be \(a\), with common difference \(d=-20\), and 7 prizes summing to 700:
So the seven prizes, each ₹20 less than the one before, are:
₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40.
Verification: \(160+140+120+100+80+60+40=700\ \checkmark\).
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant will be the same as the class in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
The number of trees planted by ONE section of Class 1, 2, 3, …, 12 is \(1,\ 2,\ 3,\ldots,\ 12\) — an AP with \(a=1,\ d=1,\ n=12\). Its sum:
So one section across all 12 classes plants 78 trees. Since there are 3 sections of every class:
So, the students will plant a total of 234 trees.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … as shown in the figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take \(\pi=\dfrac{22}{7}\))
The length of a semicircle of radius \(r\) is half the circumference of a full circle, i.e. \(\pi r\) (NOT \(2\pi r\)). So the total length of 13 consecutive semicircles is \(\pi\) times the SUM of their 13 radii:
The radii \(0.5,\ 1.0,\ 1.5,\ 2.0,\ldots\) (13 terms) form an AP with \(a=0.5,\ d=0.5,\ n=13\):
So the total length of the spiral is:
So, the total length of the spiral made up of thirteen consecutive semicircles is 143 cm.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
The number of logs per row forms an AP: \(a=20,\ d=-1\). We need \(S_{n}=200\):
By the quadratic formula, with discriminant \(41^{2}-4(400)=1681-1600=81=9^{2}\):
Both are positive integers, so both must be checked against the physical situation — the number of logs in the top (i.e. \(n\)th) row must be a POSITIVE number, since you cannot stack a negative number of logs:
For \(n=25\): \(a_{25}=21-25=-4\) — negative, impossible. Rejected.
For \(n=16\): \(a_{16}=21-16=5\) — positive, valid.
So the logs are placed in 16 rows, with 5 logs in the top row.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first and second potato, the total distance (in metres) run by a competitor is \(2\times5+2\times(5+3)\)]
To fetch the \(k\)th potato, the competitor runs from the bucket to it and back, a round trip of \(2\times\left[5+3(k-1)\right]\) metres (since the \(k\)th potato is \(5+3(k-1)\) m from the bucket). So the round-trip distances for potato 1, 2, 3, … are:
This is an AP with \(a=10\) (matching the hint’s \(2\times5\)) and \(d=6\) (each extra potato is 3 m further, so the round trip is \(2\times3=6\) m more), for \(n=10\) potatoes:
So, the total distance the competitor has to run is 370 m.
NCERT’s own note on this exercise: “These exercises are not from the examination point of view.” Included here for completeness, in the same step-by-step format as every other exercise on this page.
Which term of the AP: \(121,\ 117,\ 113,\ldots\), is its first negative term?
[Hint: Find \(n\) for \(a_{n}<0\)]
Here \(a=121,\ d=-4\). We need the smallest \(n\) for which \(a_{n}\) is negative:
The smallest integer \(n\) satisfying this is \(n=32\). Check the two terms on either side: \(a_{31}=121-4(30)=1\) (still positive), and \(a_{32}=121-4(31)=121-124=-3\) (negative). So the 32nd term is the first negative term (its value is \(-3\)).
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
\(a_{3}+a_{7}=(a+2d)+(a+6d)=2a+8d=6\ \Rightarrow\ a+4d=3\).
Let \(a=3-4d\); then \(a_{3}=a+2d=(3-4d)+2d=3-2d\) and \(a_{7}=a+6d=(3-4d)+6d=3+2d\). Their product is 8:
Both signs of \(d\) are valid (they just describe the same pair of numbers read in opposite order), so there are two possible APs:
Case \(d=\dfrac{1}{2}\): \(a=3-4\left(\dfrac{1}{2}\right)=1\).
Case \(d=-\dfrac{1}{2}\): \(a=3-4\left(-\dfrac{1}{2}\right)=5\).
So, the sum of the first 16 terms is 76 or 20, depending on which AP is meant (an increasing one starting at 1, or a decreasing one starting at 5 — both fit the given conditions equally well).
A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are \(2\dfrac{1}{2}\) m apart, what is the length of the wood required for the rungs?
[Hint: Number of rungs \(=\dfrac{250}{25}+1\)]
The rungs are 25 cm apart and the top and bottom rungs are \(2\dfrac{1}{2}\ \text{m}=250\ \text{cm}\) apart, so the number of GAPS between rungs is \(\dfrac{250}{25}=10\), which means there is ONE MORE rung than gap:
The rung lengths decrease uniformly from 45 cm to 25 cm over these 11 rungs — an AP with \(a=45,\ l=25,\ n=11\). The total length of wood needed is simply the SUM of all 11 rung lengths:
So, the length of wood required for the rungs is 385 cm (i.e. 3.85 m).
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of \(x\) such that the sum of the numbers of the houses preceding the house numbered \(x\) is equal to the sum of the numbers of the houses following it. Find this value of \(x\).
[Hint: \(S_{x-1}=S_{49}-S_{x}\)]
The house numbers preceding house \(x\) are \(1,\ 2,\ldots,\ (x-1)\), whose sum is \(S_{x-1}=\dfrac{(x-1)x}{2}\).
The house numbers following house \(x\) are \((x+1),\ldots,\ 49\), whose sum is the total \(1+2+\ldots+49\) minus the sum \(1+2+\ldots+x\):
Setting “sum before” equal to “sum after”:
Multiplying throughout by 2:
(The negative root \(x=-35\) is rejected, since a house number must be a positive integer between 1 and 49.) So, such a value of \(x\) does exist: \(x=35\).
Verification: sum of houses 1 to 34 \(=\dfrac{34\times35}{2}=595\). Sum of houses 36 to 49 \(=1225-\dfrac{35\times36}{2}=1225-630=595\ \checkmark\). Both sides equal 595.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of \(\dfrac{1}{4}\) m and a tread of \(\dfrac{1}{2}\) m (see the figure). Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build the first step \(=\dfrac{1}{4}\times\dfrac{1}{2}\times50\ \text{m}^{3}\)]
Built as a solid staircase, the block of concrete under the \(k\)th step (counting from the top down, or equivalently from the bottom up — the total is the same either way) is \(k\) times as tall as a single rise, so its volume is \(k\) times the volume of the very first step:
So the volumes of the 15 steps are \(1\times6.25,\ 2\times6.25,\ 3\times6.25,\ldots,\ 15\times6.25\ \text{m}^{3}\) — an AP with \(a=6.25,\ d=6.25,\ n=15\). The total volume is their sum:
So, the total volume of concrete required to build the terrace is 750 m³.
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