Question

How many terms are there in the AP whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?

Answer 1

Answer:

a=-14

a5=a+4d

2=-14+4d

16=4d

4=d

Sn=n/2(2a+(n-1)d)

40=n/2(2(-14)+(n-1)4)

80/n=-28+4n-4

80/n=-32+4n

80=n(-32+4n)

80=-32n+4n²

4n²-32n-80=0

n²-8n-20=0

n²+2n-10n-20=0

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

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

n=10  or  n=-2

⇒n=10

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Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

A5=a+4d=2=-14+4d d=+4

Sn=n/2(2a+(n-1)d

40=n/2(2×-14+(n-1)4

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

Equation on servlet and distributed find the term