If -i-125a1-625 such that a1-a2-a3- A-P- then -j-113a2j-1 equals

If $a_{1}$ , $a_{2}$ , $a_{3}$ , .... is an A.P. such that $a_{1}$ + $a_{5}$ + $a_{10}$ + $a_{15}$ + $a_{20}$ + $a_{24}$ = 225 then $a_{1}$ + $a_{2}$ + $a_{3}$ + ... + $a_{23}$ + $a_{24}$ is equal to

Q. If

a_{1}, a_{2}, a_{3}, . . . . a_{n}

are in A.P., where

a_{1} > 0

for all

i

, then

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}}

is equal to

Q. Let

a_{1}, a_{2}, a_{3}, .

be an

A . P .

such that

\frac{a_{1} + a_{2} + . . . . + a_{p}}{a_{1} + a_{2} + a_{3} + . . . . + a_{q}} = \frac{p^{3}}{q^{3}}; p \neq q

. Then

\frac{a_{6}}{a_{21}}

is equal to:

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If 25∑i=1a1=625 such that a1,a2,a3⋯ϵ A.P. then 13∑j=1a2j−1 equals

The correct option is C 325Given 25∑i=1ai=a1+a2+a3+....+a25=252(a1+a25)⇒625=252(a1+a25)∴a1+a25=50Now2(a1+a2+...+a12)+a13=6252×6×50+a13=625a13=25∴13∑j=1a2j−1=a1+a3+a5+...+a25=6(a1+a25)+a13=6×50+25=325

If $25 \sum i = 1 a_{1} = 625$ such that $a_{1}, a_{2}, a_{3} \dots ϵ$ A.P. then $13 \sum j = 1 a_{2 j - 1}$ equals