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If the sum of the first 15 terms of the series 3+7+14+24+37+.......is 15K, then k is equal to :

Option 1)
126

Option 2)
122

Option 3)
81

Option 4)
119

Answers (1)

S solutionqc

Use

Summation of series of natural numbers -

$\sum_{k=1}^{n}K= \frac{1}{2}n\left ( n+1 \right )$

- wherein

Sum of first n natural numbers

$1+2+3+4+------+n= \frac{n(n+1)}{2}$

and

Summation of series of natural numbers -

$\sum_{k=1}^{n}K^{2}= \frac{1}{6}n\left ( n+1 \right )\left ( 2n+1 \right )$

- wherein

Sum of squares of first n natural numbers

$1^{2}+2^{2}+3^{2}+4^{2}+------+n^{2}= \frac{n(n+1)\left ( 2n+1 \right )}{6}$

3+7+14+24+37 ----------15K

Let T_n=an²+bn+c because difference is in A.P: [4,7,10,13-----------------]

at n=1 $3=a+b+c$ ------(i)

n=2 $7=4a+2b+c$ -------(ii)

n=3 $14=9a+3b+c$ -------(iii)

(ii)-(i) $\therefore 4=3a+b$ -------- (iv)

(iii)-(ii) $7=5a+b$ -------- (v)

(v)-(iv) $3=2a$

$a=\frac{3}{2}$

from (i) $b+c=3-\frac{3}{2}=\frac{3}{2}$

from (ii) $2b+c=7-6=1$

$\therefore b=1-\frac{3}{2}=-\frac{1}{2}$

$\therefore c=1-\frac{3}{2}+\frac{1}{2}=\frac{4}{2}=2$

$\therefore T_{n}=\frac{3}{2}n^{2}-\frac{1}{2}n+2$

$S_{n}=\sum T_{n}=\frac{3}{2}\sum n^{2}-\frac{1}{2}\sum n+2\sum 1$

$=\frac{3}{2}\cdot \frac{n(n+1)(2n+1)}{6}-\frac{1}{2}\cdot \frac{n(n+1)}{2}+2n$

Put n=15 $\frac{3}{12}\times 15\times 16\times 31-\frac{1}{4}\times 15\times 16+30$

$=60\times 31-60+30$

$=60\times 30+30$

$=61\times30$

$=15K (given)$

$\therefore K=\frac{61\times 30}{15}=122$

Option 1)

126

This option is incorrect

Option 2)

122

This option is correct

Option 3)

81

This option is incorrect

Option 4)

119

This option is incorrect

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