Question

the first term of an AP is 7 and sum of its first 4 terms is half the sum of the next 4 terms. Find the sum of the first 30 terms

Answer 1

Step-by-step explanation:

Given that , First term of the A.P. = 7

Sum of first 4 terms is half the sum of the next 4 terms

=> S4 = 1/2(S8 - S4)

$=>\frac{4}{2}(2*7+(4-1)d)=\frac{1}{2}[(\frac{8}{2}(2*7+(8-1)d)-\frac{4}{2}(2*7+(4-1)d)]\\=>2(14+3d)=\frac{1}{2}[4*(14+7d)-2(14+3d)]\\=>2*(28+6d)=(56+28d-28-6d)\\=>56+12d=28+22d\\=>28=10d\\=>d=2.8$

Now,

$S_{30}=\frac{30}{2}[2*7+(30-1)*2.8]\\=15[14+81.2]\\=15*95.2\\=1428$

∴ Sum of first 30 terms of the A.P. is 1428

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Answer 2

The sum of first 30 terms is 1428.

First Term of AP = 7 (Given)

Sum of first four terms = 1/2 of next four (Given)

Thus,

S4 = 1/2(S8-S4)

= 2 ( 14+ 3d ) = 1/2 (4(14 + 7d) - 2(14+3d))

= 2 ( 28 + 6d ) = 56 + 28 d -28 -6d

= 56 + 12d  = 56 -28 +22d

= 56 - 56 -28 = 22d - 12d

= 28 = 10 d

= d= 28/10

= d = 2.8

Now, to find sum of 30 terms -  

S30 = 15 ( 14+29 × 2.8 )

= 15 [ 14 + 81.2]

= 15(95.2)

= 1428