If is an A.P. and a_1 + a_4 + a_7 + d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) In an $A.P.$,the sum of terms equidistant from the beginning and end is constant.$a_1 + a_{16} = a_4 + a_{13} = a_7 + a_{10} = \dots = \lambda$.Gi

Vedclass · Accepted Answer

$98$

Vedclass · Answer

(B) In an $A.P.$,the sum of terms equidistant from the beginning and end is constant.$a_1 + a_{16} = a_4 + a_{13} = a_7 + a_{10} = \dots = \lambda$.Given the sum $a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 147$.This can be written as $3(a_1 + a_{16}) = 147$,so $a_1 + a_{16} = 49$.We need to find $S = a_1 + a_6 + a_{11} + a_{16}$.Since $a_1 + a_{16} = a_6 + a_{11} = 49$,we have $S = (a_1 + a_{16}) + (a_6 + a_{11}) = 49 + 49 = 98$.

If $< a_n >$ is an $A.P.$ and $a_1 + a_4 + a_7 + \dots + a_{16} = 147$,then $a_1 + a_6 + a_{11} + a_{16}$ is equal to

Similar Questions

Which term of the sequence $5, 8, 11, 14, \dots$ is $320$?

If the sum of the first $2n$ terms of the arithmetic progression $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the arithmetic progression $57, 59, 61, \dots$,then $n = \dots$

The sum of all natural numbers between $1$ and $100$ which are multiples of $3$ is

If the roots of the equation $6x^3-11x^2+6x-1=0$ are in harmonic progression,then the roots of $x^3-6x^2+11x-6=0$ will be in

Let $a_1, a_2, a_3, \dots$ be an $A.P.$ such that $\frac{a_1 + a_2 + \dots + a_p}{a_1 + a_2 + \dots + a_q} = \frac{p^3}{q^3}$ where $p \neq q$. Then $\frac{a_6}{a_{21}}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module