Question

Ap whose third term is 16 and the 7th term exceed the 5 term by 12​

Answer 1

Let a be the First term, a

3

be the third term, a

5

be the 5th term and a

7

be the 7th term

We know, a

n

=a+(n−1)d

⇒ a

3

=16 [ Given ]

⇒ a+(3−1)d=16

⇒ a+2d=16 ....... (1)

⇒ Now, a

7

−a

5

=12 [ Given ]

⇒ [a+(7−1)d]−[a+(5−1)d]=12

⇒ 2d=12

∴ d=6

From equation (1), we get

a+2(6)=16

a+12=16

∴ a=4

So first term is 4

We can find AP by adding d continuously.

∴ Required AP is 4,10,16,22,28,34,40

ok my friend