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The sum of 5th and 9th terms of an AP is 30. If its 25th term is three times its 8th term, find the AP.

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+1 vote
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Best answer

a5 +a= 30

(a +4d) + (a +8d) = 30

2a +12d = 30

a + 6d = 15 (i)

Acc. To question,

a25 = 3a8

a + 24d = 3(a +7d)

3d = 2a

3d = 2(15 - 6d) [from(i)]

15d = 30

d = 2

Putting the value of d in (i),

a = 15 – 6(2) = 3

a1 = a = 3

a2 = a +d = 3 +2 = 5

a3 = a +2d = 3 + 2(2) = 7

Hence, the A.P. is 3, 5, 7, 9,.......

+1 vote
by (472 points)

HELLO,

Let the AP is \(a_1, a_2, a_3,...a_n\) with common difference d.

Given in the question,

 \(a_5 + a_9= 30\)

\(a_1+4d+a_1+8d=30\)

\(2a_1+12d=30\)

\(a_1+6d=15\)                         ------------- 1

Now again given in the question,

\(a_{25}=3a_8\)

\(a_1+24d= 3(a_1+7d)\)

\(a_1+24d= 3a_1+21d \)

\(2a_1-3d=0\)         -------------- 2

Now solving equation 1 and 2 we get     ( \(2* eqn 1 - eqn2\)),    

\(15d=30\)

d=2 and \(a_1=3\)

hence, \(a_1=3, a_2= 3+2, a_3= 3+2+2 ...\)

So, the required AP is 3,5,7,9 ...

I HOPE YOU WILL UNDERSTAND.smiley

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