Solutions: Arithmetic Progressions - Vardaan Learning Institute

1. The nth term of the AP: 5, 2, -1, -4, ... is:

Ans: (c) 8 - 3n
$a = 5, d = 2 - 5 = -3$.
$a_n = a + (n-1)d$
$= 5 + (n-1)(-3)$
$= 5 - 3n + 3$
$= 8 - 3n$.

2. If the common difference of an AP is 3, then $a_{20} - a_{15}$ is:

Ans: (c) 15
$a_{20} - a_{15} = (a+19d) - (a+14d)$ $= 5d$.
$5 \times 3 = 15$.

3. The first four terms of an AP whose first term is -2 and the common difference is -2 are:

Ans: (c) -2, -4, -6, -8
-2, -2-2=-4, -4-2=-6, -6-2=-8.

4. The sum of the first five multiples of 3 is:

Ans: (a) 45
3, 6, 9, 12, 15. Sum = 45.
Or $S_5 = 5/2 (2(3) + 4(3))$
$= 5/2 (6+12)$
$= 5/2(18)$$ = 45$.

5. If k, 2k-1 and 2k+1 are three consecutive terms of an AP, the value of k is:

Ans: (c) 3
$2(2k-1) = k + (2k+1)$.
$4k - 2 = 3k + 1 \Rightarrow k = 3$.

6. The 10th term from the end of the AP: 4, 9, 14, ..., 254 is:

Ans: (a) 209
$l = 254, d = 5$. 10th from end = $l - (10-1)d$.
$254 - 9(5) = 254 - 45 = 209$.

7. The 11th term of the AP: -5, -5/2, 0, 5/2, ... is:

Ans: (b) 20
$a = -5, d = -5/2 - (-5) = 5/2$.
$a_{11} = -5 + 10(5/2) = -5 + 25 = 20$.

8. In an AP, if $d = -4, n = 7, a_n = 4$, then a is:

Ans: (d) 28
$a_n = a + (n-1)d \Rightarrow 4 = a + 6(-4)$.
$4 = a - 24 \Rightarrow a = 28$.

9. The sum of first n positive integers is given by:

Ans: (b) $n(n+1)/2$

10. Assertion: 2, 4, 8, 16... is AP.

Ans: (d) A is false but R is true
Ratios are constant (GP), not differences. So A is False.

11. Find the sum of the first 22 terms of the AP: 8, 3, -2, ...

Ans: -979
$a=8, d=-5, n=22$.
$S_{22} = 22/2 [2(8) + 21(-5)]$
$= 11 [16 - 105]$
$= 11(-89) = -979$.

12. Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term?

Ans: 65th term
$a_n = a_{54} + 132$.
$a+(n-1)d = a+53d+132$.
$(n-1)d = 53d+132$. ($d=12$).
$12(n-1) = 53(12) + 132$.
$n-1 = 53 + 11 = 64$. $n=65$.

13. Find the number of terms in the AP: 7, 13, 19, ..., 205.

Ans: 34
$205 = 7 + (n-1)6$.
$198 = 6(n-1) \Rightarrow 33 = n-1 \Rightarrow n = 34$.

14. If the sum of the first n terms of an AP is $4n - n^2$, what is the 10th term?

Ans: -15
$a_n = S_n - S_{n-1}$.
$a_{10} = S_{10} - S_9$
$= (40-100) - (36-81)$
$= -60 - (-45) = -15$.

15. The sum of the 4th and 8th terms...

Ans: -13, -8, -3
$a_4+a_8=24 \Rightarrow 2a+10d=24$.
$a_6+a_{10}=44 \Rightarrow 2a+14d=44$.
Subtracting: $4d=20 \Rightarrow d=5$.
$2a+50=24 \Rightarrow 2a=-26 \Rightarrow a=-13$.

16. How many two-digit numbers are divisible by 3?

Ans: 30
12, 15, ..., 99.
$99 = 12 + (n-1)3$.
$87 = 3(n-1) \Rightarrow 29 = n-1 \Rightarrow n = 30$.

17. Find the sum of all odd integers between 2 and 100 which are divisible by 3.

Ans: 867
Odd multiples of 3: 3, 9, 15, ..., 99.
This is an AP with a=3, d=6, l=99.
$99 = 3 + (n-1)6$
$\Rightarrow 96=6(n-1)$
$\Rightarrow 16=n-1 \Rightarrow n=17$.
$S = 17/2 (3+99)$
$= 17/2 (102)$
$= 17 \times 51 = 867$.

18. If m times the mth term of an AP is equal to n times its nth term...

$m[a+(m-1)d] = n[a+(n-1)d]$.
$am + m^2d - md = an + n^2d - nd$.
$a(m-n) + d(m^2-n^2) - d(m-n) = 0$.
$(m-n)[a + d(m+n) - d] = 0$.
Since $m \neq n$, $a + (m+n-1)d = 0$, which is $a_{m+n}$.

19. Which term of 121, 117, 113... is first negative?

Ans: 32nd term
$a = 121, d = -4$.
$a_n < 0 \Rightarrow 121 + (n-1)(-4) < 0$.
$121 - 4n + 4 < 0 \Rightarrow 125 < 4n$.
$n > 31.25$. So $n=32$.

20. Sum of 7 terms is 49, 17 terms is 289...

Ans: $n^2$
$S_7=49 \Rightarrow 7/2(2a+6d)=49 \Rightarrow a+3d=7$.
$S_{17}=289 \Rightarrow 17/2(2a+16d)=289 \Rightarrow a+8d=17$.
Solving: $5d=10 \Rightarrow d=2, a=1$.
$S_n = n/2[2(1) + (n-1)2] = n/2[2+2n-2] = n/2[2n] = n^2$.

21. Spiral length...

Ans: 143 cm
Semi-circumference = $\pi r$.
$l_1 = \pi(0.5)$, $l_2 = \pi(1.0)$, ...
AP with $a=\pi/2, d=\pi/2$. Total length $S_{13}$.
$S_{13} = 13/2 [2(\pi/2) + 12(\pi/2)]$
$= 13/2 [14\pi/2] = 13/2 [7\pi]$
$= 13/2 [7(22/7)] = 13 \times 11 = 143$.

22. Potato race distance...

Ans: 370 m
Distances run: $2(5)$, $2(5+3)$, $2(5+6)$...
10 terms. $a=10, d=6$.
$S_{10} = 10/2 [2(10) + 9(6)]$
$= 5 [20 + 54] = 5(74) = 370$ m.

23. Case Study: Savings Plan

a=100, d=50.
(i) $a_{10} = 100 + 9(50) = 100 + 450 = ₹550$.
(ii) $S_{10} = 10/2 [200 + 9(50)]$
$= 5 [200 + 450]$
$= 5(650) = ₹3250$.
(iii) $500 = 100 + (n-1)50$
$\Rightarrow 400 = 50(n-1)$
$\Rightarrow 8 = n-1 \Rightarrow n=9$. 9th month.