Question

the sum of n term of an a.p. is 136 and common deference is 4. if the last term is 31, then find the number of terms.​

Answer 1

Answer:

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Answer 2

Answer:

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Step-by-step explanation:

The number of terms is 8.

Step-by-step explanation:

Since we have given that

Sum of n terms of an A.P. = 136

Common difference (d) = 4

Last term = 31

As we know the formula for "Last term":

\begin{lgathered}a_n=a+(n-1)d\\\\31=a+(n-1)4\\\\31=a+4n-4\\\\31+4=a+4n\\\\35=a+4n\end{lgathered}

a

n

=a+(n−1)d

31=a+(n−1)4

31=a+4n−4

31+4=a+4n

35=a+4n

Similarly, we know the formula for "Sum of n terms ":

\begin{lgathered}S_n=\frac{n}{2}(2a+(n-1)d)\\\\136\times 2=n(2a+(n-1)4)\\\\372=n(2a+4n-4)\\\\\text{ using the value in eq (1)}\\\\372=n(2(35-4n)+4n-4)\\\\372=n(70-8n+4n-4)\\\\372=n(70-4n-4)\\\\372=n(66-4n)\\\\4n^2-66n+372=0\\\\2n^2-33n+136=0\end{lgathered}

S

n

=

2

n

(2a+(n−1)d)

136×2=n(2a+(n−1)4)

372=n(2a+4n−4)

using the value in eq (1)

372=n(2(35−4n)+4n−4)

372=n(70−8n+4n−4)

372=n(70−4n−4)

372=n(66−4n)

4n

2

−66n+372=0

2n

2

−33n+136=0

now, by using the "quadratic formula ", we get that

n=8,n=\frac{17}{2}n=8,n=

2

17

Since the value of n can't be in fraction or decimal.

Hence, the number of terms is 8.