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1 week, 4 days ago

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

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)
72 Views
S solutionqc
Answered 1 week, 4 days ago

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 Tn=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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