how many terms of an AP must be taken for their sum to be equal to 120 if its third term is 9 and the difference between the seventh and second term is 20

T

_{nth}= a + (n-1)d

T

_{3rd}

_{ }= 9 = a + (3-1)d

⇒a+2d = 9 (i)

and T

_{7th}-T

_{2d}= {a +(7-1)d }-{a+(2-1)d} = 20

⇒ {a+6d}-{a+d}=20

⇒5d = 20

⇒d =20/5 = 4

Substituting the value of d in equation (i), we get a + 2тип4 = 9, a = 1.

Sum upto nth term is given by S

_{nth}= n/2{2a+(n-1)d}

So Snth = 120 = n/2{2(1) + (n-1)4}

⇒ n{2+4n-4} = 240

⇒n{4n-2} = 240

4n

^{2}-2n-240 = 0

⇒ 2n

^{2}

^{ }-n -120 =0

⇒2n

^{2}

^{ }-16n+15n -120 = 0

⇒ 2n(n-8)+15(n-8) = 0

⇒ (2n+15)(n-8) = 0

n = -15/2, 8 so taking n = 8, as n can never be fraction or negative.

So sum of 8 terms of the AP =120.

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