Question

Q20) If How many terms of the 1
A.P. -6,11/2,-5,....are needed to
give the sum -25 ?
*​

Answer 1

Answer:

5 terms

Step-by-step explanation:

AP_ -6, -11/2, -5, ..

a= -6

D= -1/2

S= -25

S= n/2[2a+(n-1)d]

-25 = n/2[ 2(-6) + (n-1)-1/2]

after solving for n

n = 5

5 terms are needed to give the sum -25

Answer 2

Answer:

Given:

$\[\begin{align} & d=-\frac{11}{2}-(-6) \\ & d=\frac{1}{2} \\ \end{align}\]$$d=-\frac{11}{2}-(-6) \\ d=\frac{1}{2}$

To find: Number of terms to get a sum of -25

Solution:

$& {{S}_{n}}=\frac{n}{2}\left[ 2a+(n-1)d \right] \\ & -25=\frac{n}{2}\left[ 2(-6)+(n-1)\frac{1}{2} \right] \\ & -50=n\left[ -12+\frac{n}{2}-\frac{1}{2} \right] \\ & -100={{n}^{2}}-25n \\ & ={{n}^{2}}-25n-100 \\ & =(n-20)(n-5) \\ & n=20,5$

We'll take n=5