Question

Explain with step -by- step ​

Answer 1

Given :

• A.P.= 16, 14, 12...
• Sum of n terms = 60

To Find :

The value of n.

Solution :

Analysis :

Here we have to use the required formula. Then from there forming a quadratic equation we can find the value(s) of n.

Required Formula :

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

where,

• Sₙ = Sum of respective term
• n = respective term
• a = first term
• d = common difference

Explanation :

We know that if we are given the first term, common difference, the sum and is asked to find the number of terms then our required formula is,

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

where,

• Sₙ = 60
• n = n
• a = 16
• d = -2

Using the required formula and Substuting the required values,

$\\ :\implies\sf60=\dfrac{n}{2}[2(16)+(n-1)-2]$

$\\ :\implies\sf60=\dfrac{n}{2}[2\times16+(n-1)-2]$

$\\ :\implies\sf60=\dfrac{n}{2}[32+(-2n)-(-2\times1)]$

$\\ :\implies\sf60=\dfrac{n}{2}[32+(-2n)-(-2)]$

$\\ :\implies\sf60=\dfrac{n}{2}[32-2n+2]$

$\\ :\implies\sf60=\dfrac{n}{2}[32+2-2n]$

$\\ :\implies\sf60=\dfrac{n}{2}[34-2n]$

$\\ :\implies\sf60=\left(\dfrac{n}{2}\times34\right)-\left(\dfrac{n}{2}\times2n\right)$

$\\ :\implies\sf60=\dfrac{n}{\not{2}}\times\cancel{34}\ \ ^{17}-\dfrac{n}{\not{2}}\times\not{2}n$

$\\ :\implies\sf60=n\times17-n\times n$

$\\ :\implies\sf60=17n-n^2$

$\\ :\implies\sf n^2-17n+60=0$

Splitting the middle term,

$\\ :\implies\sf n^2-12n-5n+60=0$

$\\ :\implies\sf n(n-12)-5(n-12)=0$

$\\ :\implies\sf(n-5)(n-12)=0$

$\\ :\implies\sf(n-5)=0$

$:\implies\sf n-5=0$

$:\implies\sf n=5$

$\therefore\boxed{\bf n=5.}$

$\\ :\implies\sf(n-12)=0$

$:\implies\sf n-12=0$

$:\implies\sf n=12$

$\therefore\boxed{\bf n=12.}$

The value of n is 5 or 12.

We get values of n = 5 , 12.In this A.P.., 9th term becomes zero. The terms before 9th term are positive. The terms after 9th term are negative. By adding these negative terms to positive terms from 6th to 8th terms, [16, 14 , 12 , 10 , 8 , 6 , 4 , 2 , 0 , -2 , -4 , -6] they cancel out each other and then the remaining terms are 16 , 14 , 12 , 10 , 6. So, the sum remains same.