The sum of first 16 terms of
Question:

The sum of first 16 terms of the AP 10, 6, 2, … is

(a)-320                     

(b) 320                      

(c)-352                     

(d)-400

Solution:

(a) Given, AP is 10, 6, 2,…

$\therefore$ $S_{16}=\frac{16}{2}[2 a+(16-1) d]$ $\left[\because S_{n}=\frac{n}{2}\{2 a+(n-1) d\}\right]$

$=8[2 \times 10+15(-4)]$

 

$=8(20-60)=8(-40)=-320$

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