Question

(ii) How many terms are there in the A.P. whose first and fifth terms are – 14 and 2
respectively and the sum of the terms is 40 ?
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Answer 1

Answer:

We have,

Let first term is an A.P.

a=−14

Fifth term is a

5

=a+4d=2

Put a=−14 and we get,

a+4d=2

−14+4d=2

4d=2+14

4d=16

d=4

Find the value of n

Whose sum is 40.

Then,

We know that,

S /n= n/2(2a+(n−1)d)⇒40= n/2 (2×(−14)+(n−1)×4)

⇒80=n(−28+4n−4)

⇒80=n(−32+4n)

⇒4n(n−8)=80

⇒n(n−8)=20

⇒(n ) 2 −8n−20=0

⇒(n ) 2−(10−2)n−20=0

⇒(n ) 2 −10n+2n−20=0

⇒n(n−10)+2(n−10)=0

⇒(n−10)(n+2)=0

For

n+2=0

n=−2

It is not possible (negative)

For,

n−10=0

n=10