Question

find the sum of first 21 terms of an A.P whose 2nd term is 8 and 4th term is 14

Answer 1

may my ans will help uh..

Answer 2

735 is the sum of first 21 terms whose second term is 8 and fourth term is 14.

Given:

a + d = 8

a + 3d = 14

To find:

$S_{21}=?$

Solution:

a + d = 8 …....(1)

a + 3d = 14..…(2)

Subtracting eq(2) and (1), we get,  

2d = 6

d = 3

substitute in eq(1)

a + 3 = 8

a = 5

Hence the 21st term is

$\bold{t_{n}=a+(n-1) d}$

$\begin{array}{l}{t_{21}=5+(21-1) 3} \\ {t_{21}=5+(20) 3} \\ {t_{21}=65}\end{array}$

Hence the last term in the series (l) is  = 65

$\begin{array}{l}{S_{n}=\frac{n}{2}[a+l]} \\ {S_{21}=\frac{21}{2}[5+65]} \\ {S_{21}=\frac{21}{2}[70]}\end{array}$

$\begin{array}{l}{S_{21}=21 \times 35} \\ {S_{21}=735}\end{array}$

The sum of first 21 terms is 735.