Question

In an AP, the sum of Ist n terms is 3n^2/2+13n/2. Find its 25th term and common difference.​

Answer 1

Answer :-

$\sf{\implies \;Value\: of\: 25th \:term \:of \:this\: AP\: is \:80 }$

SOLUTION ;-

Given that sum of 1st n terms of AP is $\sf{\implies \frac{3n^{2} }{2} + \frac{13n}{2} }$

WE know that sum of 1st 25 terms - sum of 1st 24 terms = 25th term of the given AP .

Firstly we'll find the value of sum of 1st 25 terms of AP .

Putting n as 25

$\sf{\implies \frac{3(25)^{2} }{2} + \frac{13(25)}{2} } \;\;\;\;\;\;\ \boxed{eq \; 1st }$

Now putting n as 24

$\sf{\implies \frac{3(24)^{2} }{2} + \frac{13(24)}{2} } \;\;\;\;\;\;\boxed{eq \; 2nd}$

Now substract eq 2nd from eq 1st :-

$\sf{\implies \frac{3({25)}^{2}}{2} + \frac{13(25)}{2} - [ \frac{3(24)^{2} }{2} + \frac{13(24)}{2} ]}$

$\sf{\implies [\frac{3(25)^{2} }{2} - \frac{3(24)^{2} }{2} ]+ [\frac{13(25)}{2} - \frac{13(24)}{2}]}$

Taking $\frac{3}{2}$ common from 1sr bracket and $\frac{13}{2}$ common from 2nd bracket .

$\sf{\implies (\frac{3}{2} )(625 - 576 ) + (\frac{13}{2} )(25 - 24 ) }$

$\sf{\implies \frac{3 \times 49 }{2} + \frac{13}{2}}$

$\sf {\implies \frac{147}{2} + \frac{13}{2}}$

$\sf {\implies \frac{147 + 13}{2} = \frac{160}{2} }$

$\underline{\underline{\sf{\implies 80 \; is\;the\: answer}}}$

Other method :-

Putting n = 1

$\sf{\implies \frac{3({1}^{2}) }{2} + \frac{13(1)}{2} }$

$\sf{\implies a_1 \; = \frac{3}{2} + \frac{13}{2}}$

$\sf{\implies a_1 = \frac{3 + 13}{2} }$

$\sf{\implies a_1 = \frac{16}{2} }$

$\underline{\sf{\implies a_1 = 8}}$

Now by putting n equals 2 we got the sum of first two terms of AP.

$\sf{\implies \frac{3 ({2}^{2})}{2} + \frac{13(2)}{2}}$

$\sf{\implies \frac{3(4)}{2} + 13 }$

$\sf{\implies 6 + 13 = 19 }$

2nd term of AP = 19 - 8 = 11 .

Common difference = 11 - 8

→ Common difference = 3

$\sf{\implies 25th\: term \: = a+24(d) }$

$\sf{\implies 8 + 24(3) }$

$\sf{\implies 8 + 72 = 80}$

$\underline{\underline{\sf{\implies Hence \:,\: 80 \; is\;the\: answer}}}$