Question

10+12+14+...+(2n)=90 n=?

Answer 1

Answer:

n=10

Step-by-step explanation:

The given equation is  

$10+12+14+...+2n=90$        ...(1)

Let nth term is 2n.

We need to find the values of n.

It is clear that $S_n=10+12+14+...+2n$ is sum of A.P., whose first term is 10 and common difference is 2.

Sum of A.P. is  

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

where, a is first term and d is common difference.

Substitute a=10 and d=2 in the above formula.

$S_n=\dfrac{n}{2}[2(10)+(n-1)2]$

$10+12+14+...+2n=n[10+n-1]$

$10+12+14+...+2n=n[9+n]$     ...(2)

From (1) and (2), we get

$n(9+n)=90$

$n^2+9n-90=0$

Splitting middle term, we get

$n^2+15n-6n-90=0$

$n(n+15)-6(n+15)=0$

$(n+15)(n-6)=0$

$n=-15,6$

Since, n represents the number of terms so n cannot be a negative number, therefore number of term is 6.

Note: nth term and variable n both are different.

$a_n=a+(n-1)d$

$a_6=10+(6-1)2=10+10=20$

Sixth term is 20. So,

$2n=20$

$n=10$

Therefore, the value of n is 10.