Question

3. In an arithmetic sequence, the sum of
first 15 term is 495 and sum of first 25
terms is 1325. Find the sum of the first 'n'
terms.​

Answer 1

GIVEN :-

• Sum of first 15 terms of an arithmetic sequence , $\sf S_{15}$ = 495

• Sum of first 25 terms of an arithmetic sequence , $\sf S_{25}$ = 1325

TO FIND :-

• Sum of first n terms of the arithmetic sequence .

SOLUTION :-

    $\sf S_n = \dfrac{n}{2}\times ( 2a+(n-1)d )$

• $\sf S_{15}=\dfrac{15}{2} \times (2a+(15-1)d)$

              $\longrightarrow\sf \dfrac{15}{2}\times(2a+14d)=495\\\\\longrightarrow 15\times(2a+14d)=990\\\\\longrightarrow2a+14d=66\\\\\longrightarrow a+7d = 33 \ \ \ \ \ \ \ \ \ \ \ \ \ \....................(1)$

• $\sf S_{25} =\dfrac{25}{2} \times (2a+(25-1)d)$

             $\longrightarrow\sf \dfrac{25}{2}\times(2a+24d)=1325 \\\\\longrightarrow 25\times (2a+24d)=2650 \\\\\longrightarrow 2a+24d = 106 \\\\\longrightarrow a+12d = 53 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .......................(2)$

   Eq (2) - Eq (1) ,

     $\longrightarrow\sf 5d = 20\\\\\longrightarrow d= 4$

 Substitute the value of y in Eq (1) ,

    $\longrightarrow\sf a+7\times4 = 33\\\\\longrightarrow a=33-28\\\\\longrightarrow a =5$

 $\longrightarrow\bf Sum \ of \ n \ terms = \dfrac{n}{2} \times{(2\times5 +(n-1)4 )$

                                     $\bf = \dfrac{n}{2}\times(10+4n-4)\\\\=\dfrac{n}{2}\times (6+4n)\\\\=\dfrac{6n+4n^2}{2}\\\\\underline{\underline{=2n^2+3n}}$

Answer 2

Answer:

See on gloogle

Step-by-step explanation: