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Difference of Two Sets

If b1,b2,b3...

Question

If $b_{1}, b_{2}, b_{3} . . .$ form an A.P such that $b_{1} + b_{4} + b_{7} + . . . . + b_{28} = 220$ , then the value of $b_{1} + b_{2} + b_{3} . . . + . . . . + b_{28}$ equals

616

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308

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2200

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1232

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Solution

The correct option is A $616$
Given $b_{1} + b_{4} + b_{7} + . . . + b_{25} + b_{28} = 220$
Now $1, 4, 7, 10, . . .25, 28 ϵ A . P .$

$∴ n = 10$
and $b_{1} + b_{28} = b_{4} + b_{25} = b_{7} + b_{22} = . . . . . = b_{13} + b_{16}$
$∴ 5 (b_{1} + b_{28}) = 220 \Rightarrow b_{1} + b_{28} = 44$
Now $b_{1} + b_{2} + b_{3} + . . . + b_{28}$
$= 14 (b_{1} + b_{28}) = 14 \times 44 = 616$
Ans: A

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