Question

In an AP
(i) given a=5,d=3,$a_n$=50, find n and $S_n$.
(ii) given $a_{12}$=37,d=3, find a and $S_{12}$.
(iii) given d=5,$S_9$=75, find a and $a_9$.
(iv) given a=8,$a_n$=62,$S_n$=210, find n and d.
(v) given l=28,S=144, and there are total of 9 terms. Find a.

Answer 1

(i) Given a=5, d=3, $a_n$=50, find n and $S_n$.

Using formula $a_n$=a+ (n−1) d, to find $n^{th}$ term of arithmetic progression, we can say that

$\Rightarrow$$a_n$=5+ (n−1)(3)

$\Rightarrow$50=5+3n−3

$\Rightarrow$48=3n

$\Rightarrow$n=$\frac{48}{3}$=16

Applying formula, $S_n$=$\frac{n}{2}$(2a+ (n−1)d) to find sum of n terms of AP, we get

$S_{16}$=16/2(10+ (16−1) (3)) =8(10+45) =8(55) =440

Therefore, n=16 and $S_n$=440

(ii) Given $a_{12}$=37, d=3, find a and $S_{12}$.

Using formula $a_n$=a+ (n−1) d, to find $n^{th}$ term of arithmetic progression, we can say that

$a_{12}$=a + (12−1) 3

$\Rightarrow$37=a+33

$\Rightarrow$a=4

Applying formula, $S_n$=$\frac{n}{2}$(2a + (n−1) d) to find sum of n terms of AP, we get

$S_{12}$=$\frac{12}{2}$(8+ (12−1) (3) =6(8+33) =6(41) =246

Therefore, a=4 and $S_{12}$=246

(iii) Given d=5, $S_9$=75, find a and $a_9$.

Applying formula, $S_n$=$\frac{n}{2}$ (2a+ (n−1) d) to find sum of n terms of AP, we get

$S_9$=$\frac{9}{2}$ (2a+ (9−1) 5)

$\Rightarrow$75=$\frac{9}{2}$ (2a+40)

$\Rightarrow$150=18a+360

$\Rightarrow$−210=18a

$\Rightarrow$a=−$\frac{210}{18}$ =−$\frac{35}{3}$

Using formula $a_n$=a+ (n−1) d, to find $n^{th}$ term of arithmetic progression, we can say that

$a_9$=−$\frac{35}{3}$ + (9−1) (5) =−$\frac{35}{3}$ +40 = $\frac{(-35+120)}{3}$ = $\frac{85}{3}$

Therefore, a=$\frac{-35}{3}$ and $a_9$=$\frac{85}{3}$

(iv) Given a=8, $a_n$=62, $S_n$=210, find n and d.

Using formula $a_n$=a+ (n−1) d, to find $n^{th}$ term of arithmetic progression, we can say that

62=8+ (n−1) (d) =8+nd−d

$\Rightarrow$62=8+nd−d

$\Rightarrow$nd−d=54

$\Rightarrow$nd=54+d … (1)

Applying formula, $S_n$=$\frac{n}{2}$ (2a+ (n−1) d) to find sum of n terms of AP, we get

210=$\frac{n}{2}$ (16+ (n−1) d) =$\frac{n}{2}$ (16+nd−d)

Putting equation (1) in the above equation, we get

210=$\frac{n}{2}$ (16+54+d−d) =$\frac{n}{2}$ (70)

$\Rightarrow$$\frac{210}{70}$ =$\frac{n}{2}$

$\Rightarrow$n=6

Putting value of n in equation (1), we get

$\Rightarrow$6d=54+d

$\Rightarrow$d=$\frac{54}{5}$

Therefore, n=6 and d=$\frac{54}{5}$

(v) Given l=28, S=144, and there are total of 9 terms. Find a.

Applying formula, $S_n$=$\frac{n}{2}$ [a +l], to find sum of n terms, we get

144=$\frac{9}{2}$ [a+28]

$\Rightarrow$$\frac{288}{9}$ =[a+28]

$\Rightarrow$32=a+28

$\Rightarrow$a=4