Question

plesee ans the above que.to touch it

Answer 1

Hey there!!

Here's your answer...

Let us assume the A.P to be a, a+d, a+2d...

Second term is a2 = a + d

Seventh term is a7 = a + 6d

Fifteenth term is a15 = a + 14d

Eighth term is a8 = a + 7d

Sum of a2 and a7 is 30

So, a + d + a + 6d = 30

2a + 7d = 30

Also, a15 is 1 less than twice of a8

So, a + 14d = 2a + 14d - 1

⇒ a = 1

Since, 2a + 7d = 30

2 + 7d = 30 ⇒ 7d = 28 ⇒ d = 4

So, the A.P is 1,5,9,13...


Hope it helps!!

Answer 2

We know that sum of n terms of an AP an = a + (n - 1) * d.

Given that sum of 2nd and 7th terms of an AP is 30.

= > (a + d) + (a + 6d) = 30

= > 2a + 7d = 30   ---------- (1)


Given that its 15th term is 1 less than twice its 8th term.

= > a + 14d = 2(a + 7d)  - 1

= > a + 14d = 2a + 14d - 1

= > a - 2a = -1

= > a = 1.


Substitute a = 1 in (1), we get

= > 2a + 7d = 30

= > 2(1) + 7d = 30

= > 2 + 7d = 30

= > 7d = 28

= > d = 4.


Therefore the AP is 1,5,9....




Hope this helps!