Question

Which term of the A.P. 6, 13, 20, 27 . . . . is 98 more than its 24th term?

Answer it fast plss..⚡

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Answer 1

Answer:

The 24.th trm of the se.quence

6,13,20, 27...

6,13,20, 27...t24 = 6 + (24 – 1) 7

6,13,20, 27...t24 = 6 + (24 – 1) 7= 6 + 23 × 7 = 6 +161 =167

le.t the req.uired nth trm = 265

= 265 265 = 6 + ( n –1) 7

= 265 265 = 6 + ( n –1) 7 (n–1) 7 = 265 – 6 = 259

= 265 265 = 6 + ( n –1) 7 (n–1) 7 = 265 – 6 = 259 n – 1 =259/7= 37

7= 37n=38

hope it helps u

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Answer 2

Answer:

$\fbox{Solve :-}$

Given AP is 6,13,20,27...

First term a = 6

Common difference d = t2 - t1 = 13 - 6 = 7

nth term = tn = a + (n-1) d

Let, pth term = t24 + 98

a + (p-1) 7 = a + (24-1) 7 + 98

(p-1) 7 = 23 × 7 + 98

(p-1) 7 = 7(23+14)

p-1 = 37

p = 38

Required term = 38

Hope it helps you :)