How many terms of the A.P. -6, $$- \dfrac{11}{2}$$, -5, …. are needed to give the sum –25?

Asked by Pragya Singh | 1 year ago |  104

##### Solution :-

-25 is the sum of n terms of the given A.P.

The common difference of the A.P. is $$d=-\dfrac{11}{2}+6$$

$$\dfrac{-11+12}{2}= \dfrac{1}{2}$$

The sum of first n terms of an arithmetic progression is given by the equation

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

Substitute Sn=-25, a=-6 and $$d= \dfrac{1}{2}$$ in the equation

$$-25= \dfrac{n}{2}[2(-6)+(n-1)\dfrac{1}{2}]$$

$$-50= n[-12+\dfrac{n}{2}-\dfrac{1}{2}]$$

$$-50= n[\dfrac{25}{2}+\dfrac{1}{2}]$$

-100 = n [-25 + n]

= n2-25n+100=0

Factorize the equation.

= n2-5n+20n+100=0

= n(n-5)-20(n-5)=0

= n=20 or 5

Therefore, 5 or 20 terms of the A.P. are needed to give the sum -25  .

Answered by Abhisek | 1 year ago

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