JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 1)
Let a1, a2, a3, . . . . . . . , an, . . . . . be in A.P.
If a3 + a7 + a11 + a15 = 72,
then the sum of its first 17 terms is equal to :
If a3 + a7 + a11 + a15 = 72,
then the sum of its first 17 terms is equal to :
306
153
612
204
Explanation
As a1 a2 . . . . . an . . . . . are in A.P.
$$ \therefore $$ a3 + a15 = a7 + a11 = a1 + a17
Given,
a3 + a7 + a11 + a15 + a15 = 72
$$ \Rightarrow $$ (a3 + a15) + (a7 + a11) = 72
$$ \Rightarrow $$ 2(a1 + a17) = 72
$$ \Rightarrow $$ (a1 + a17) = 36
$$ \therefore $$ Sum of first 17 terms
= $${{17} \over 2}$$ (a1 + a17)
= $${{17} \over 2}$$ $$ \times $$ 36
= 306
$$ \therefore $$ a3 + a15 = a7 + a11 = a1 + a17
Given,
a3 + a7 + a11 + a15 + a15 = 72
$$ \Rightarrow $$ (a3 + a15) + (a7 + a11) = 72
$$ \Rightarrow $$ 2(a1 + a17) = 72
$$ \Rightarrow $$ (a1 + a17) = 36
$$ \therefore $$ Sum of first 17 terms
= $${{17} \over 2}$$ (a1 + a17)
= $${{17} \over 2}$$ $$ \times $$ 36
= 306
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