Question

the sum of the 2nd and 7th terms of an A.P. is 30. if its 15th term is 1 less than twice its 8th term, find the A.P.

Answer 1

Let the first term a and common difference be d.a2+a7 = 30a+d + a+6d = 302a + 7d = 30−−−−−−−−−−−−−−−−−(1)a15 = 2a8 − 1a+14d = 2(a+7d) − 1a+14d = 2a  + 14d −1a = 1Using (1),2+7d=307d=28d=4So, the AP will be:1, 5, 9, 13, 17.....

OR

Dear Student!

Given, sum of first n terms of an AP, Sn = 3n2 – n

 Sn = 3n2 – n

Replacing n by n – 1, we get


Let the nth term of AP be an.

Now,put n = 1 in an, we get a = 6×1−4 = 2put n = 2 in an, we get a2 = 6×2−4 = 8put n = 3 in an, we get a3 = 6×3−4 = 14and so on.So, required AP is, 2,8,14,.....

Putting n = 25, we get

a25 = 6 × 25 – 4 = 150 – 4 = 146

Thus, the 25th term of the given A.P. is 146.

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Answer 2

Here is your solution

Given:-

t2+t7=30
=>a+d+a+6d=30
=>2a+7d=30..................... (1)

A/q
15th term=2×8th term-1
=>a+14d=2×(a+7d)-1
=>a+14d=2a+14d-1
=>14d-14d+1=2a-a
=>1=a

Putting the value of a in equation (1)

2a×7d=30
2×1+7d=30
2+7d=30
7d=30-2
7d=28
d=4

A.p are a,a+d,a+2d,a+3d,a+4d

=>A.p are 1,5,9,13.........

HOPE IT HELPS YOU