Question

(b) The sum of the first 13 terms of an arithmetic progression is 312 and the sum of the next 13 terms is 819. Find the first term and common difference of this arithmetic progression. Hence, find the S40​

Answer 1

Answer:

Step-by-step explanation:

$\displaystyle u_1+u_2+u_3+...+u_{13}=\sum_{i=1}^{13}u_i=\dfrac{(u_1+u_{13})*13}{2} \\=\dfrac{(u_1+u_1+12*d)*13}{2} \\=(u_1+6d)*13=312\\\\u_1+6d=24\ (1) \\\\----------------------------\\\displaystyle u_{14}+u_{15}+u_{16}+...+u_{26}=\sum_{i=14}^{26}u_i=\dfrac{(u_{14}+u_{26})*13}{2} \\=\dfrac{(u_1+13*d+u_1+25*d)*13}{2} \\=(u_1+19d)*13=819\\\\u_1+19d=63\ (2)\\\\(2)-(1) ==> 13d=39 ==> d=3\\\\u_1=24-6*3=6\\\\S_{40}=2580$