Let a 1 , a 2 , a 3 , … , a n , … be in A.P. If a 3 + a 7 + a 11 + a 15 = 72 , then the sum of its first 17 terms is equal to

Question

Let $a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots$ be in A.P. If $a_{3} + a_{7} + a_{11} + a_{15} = 72,$ then the sum of its first 17 terms is equal to _______.

Smart Achievers · Accepted Answer

(306)

$We have, a_{3} + a_{7} + a_{11} + a_{15} = 72$

$\Rightarrow$ $(a_{3} + a_{15}) + (a_{7} + a_{11}) = 72$

$Now, a_{3} + a_{15} = a_{7} + a_{11} = a_{1} + a_{17}$ $∴$ $a_{1} + a_{17} = 36$

$Now, S_{17} = \frac{17}{2} [a_{1} + a_{17}] = 17 \times 18 = 306$

Solution

Q.
Let $a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots$ be in A.P. If $a_{3} + a_{7} + a_{11} + a_{15} = 72,$ then the sum of its first 17 terms is equal to _______.

Ans.
(306)

$We have, a_{3} + a_{7} + a_{11} + a_{15} = 72$

$\Rightarrow$ $(a_{3} + a_{15}) + (a_{7} + a_{11}) = 72$

$Now, a_{3} + a_{15} = a_{7} + a_{11} = a_{1} + a_{17}$ $∴$ $a_{1} + a_{17} = 36$

$Now, S_{17} = \frac{17}{2} [a_{1} + a_{17}] = 17 \times 18 = 306$

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Solution

Q. 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 _______.

Ans. (306) We have, a3+a7+a11+a15=72 ⇒ (a3+a15)+(a7+a11)=72 Now, a3+a15=a7+a11=a1+a17 ∴ a1+a17=36 Now, S17=172[a1+a17]=17×18=306

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Q.
Let $a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots$ be in A.P. If $a_{3} + a_{7} + a_{11} + a_{15} = 72,$ then the sum of its first 17 terms is equal to _______.

Ans.
(306)

$We have, a_{3} + a_{7} + a_{11} + a_{15} = 72$

$\Rightarrow$ $(a_{3} + a_{15}) + (a_{7} + a_{11}) = 72$

$Now, a_{3} + a_{15} = a_{7} + a_{11} = a_{1} + a_{17}$ $∴$ $a_{1} + a_{17} = 36$

$Now, S_{17} = \frac{17}{2} [a_{1} + a_{17}] = 17 \times 18 = 306$