Question

Which term of the A.P. 3, 14, 25, 36, … will be 77 more than its 30th term? Find the term.​

Answer 1

Step-by-step explanation:

n AP, the first term = a = 3

Common difference = d = 14 – 3 = 11

To find: place of the term which is 99 more than its 25th term.

So, we first find its 25th term.

Since, we know that

an = a + (n-1) × d

∴ a25 = 3 + (25 -1) × 11

⇒ a25 = 3 + 24 × 11

⇒ a25 = 3 + 264

⇒ a25 = 267

∴ 25th term of the AP is 267.

Now, 99 more than 25th term of the AP is 99 + 267 = 366.

So, to find: place of the term 366.

So, let an = 366

Since, we know that

an = a + (n-1) × d

∴ 366 = 3 + (n -1) × 11

⇒ 366 – 3 = 11n - 11

⇒ 363 = 11n - 11

⇒ 363 + 11 = 11n

⇒ 11n = 374

⇒ n = 374/11 = 34

∴ 34th term of the AP is the term which is 99 more than 25th term.

Answer 2

d=14-3=11

a=3

un = u1+(n-1)d

u30 =3+29*11 = 322 >77