Question

find the sum of the first 22terms of AP whose second and fourth term is 32 and28 respectively​

Answer 1

Answer:

1034

Step-by-step explanation:

28=a+d

32=a+3d

Solving both equation:

4=2d

d=2

Then,

28=a+d

28=a+2

a=26

Again,

First term=a+d

=26+2

=28

Sum of 22th terms=n/2[2a+d(n-1)]

=22/2[2*26+2(22-1)]

=11[52+42]

=11*94

=1034. ans.

Answer 2

Answer:

here, t2 =32, t4=28,n=22 [Given]

tn= a+(n-1)d [Formula]

t2=a+(n-1)d

but t2=32,n=22 [ given]

.

. . 32=a+(22-1)d

32=a+21d (1)

Now, t4=a+(n-1)d

but t4=28,n=22 [given]

.

. .28=a+(22-1)d

28=a+21d (2)

Adding (1) and (2)

a+21d=32

+ a+21d=28

____________

2a+42d=60

dividing both sides by 2 we get,

a+21d=30 (3)

Now,Sn=n/2(2a+(n-1)d

S22=22/2(2a+(22-1)d

=22×a+21d

=22×30 [from (3)]

=660