Question

The sum of the first three terms of an A.P. is $9$ and the sum of their square is $35$. The sum to first $n$ terms of the series can be

• A. $n(n+1)$
• B. $n^2$
• C. $n(4-n)$
• D. $n(6-n)$

Answer 1

Answer: (B), (D)

The correct options are
B $n^2$
D $n(6-n)$

Sum of first three terms of AP = 9

Sum of square of three terms of AP=35

Let AP is $a,a+d,a+2d$........

with first terms = a

common different = d

$\to a+a+d+a+2d=9$

$3a+3d=9$

$a+d=3$

$\to a^2+(a+d{)}^2+(a+2d{)}^2=35$

Put $d=3-a$

$a^2+3^2+(6-a{)}^2=35$

$2a^2+9+36-12a=35$

$2a^2-12a+10=0$

$a^2-6a+5=0$

$(a-1)(a-5)=0$

$a=1,5$

That given $d=2,-2$

Sum of first n term $=\frac{n}{2}(2a+(n-1\left)d\right)$

$\frac{n}{2}(2+2(n-1)=n^2[a=1,d=2]$

Or

$\frac{n}{2}[10-2(n-1\left)\right]$

$n[5-n+1]$

$n[6-n]$

$B,D$ are correct