Question

How many terms are there in AP whoes first term is -14 and common difference is 4 and the sum of term is 40

Answer 1

Given , a = -14

            d = 4

Let there be 'n' terms

and Sn = 40

n/2[2a + (n-1)d] = 40

so, n[-28 + (n-1)4] = 80

     n[-28 +4n - 4] = 80

     n[4n-32] = 80

     n[n-8] = 20

     n^2 - 8n -20 = 0

Hence, n = 10 or n = -2

Since, n can' be negative.

Thus, There are 10 terms in AP.

Answer 2

HEY MATE♡

HERE IS YOUR ANSWER

a= -14

d=4

sn=40

n=?

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

40×2=n[-28+(n-1)4]

80=n(-28+4n-4)

80=n(-32+4n)

80=-32n+4n^2

0=4n^2-32n-80

0=4(1n^-8n-20)

0=n^2-8n-20

0=n^2-10n+2n-20

0=n(n-10)+2(n-10)

0=(n+2)(n-10)

n+2=0 OR n-10=0

n= -2 OR n=10

As number of terms cannot be negative so, n=10

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HOPE IT HELPED ^_^