How many terms are there in the A.P. whose first and fifth terms are -

1.	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?
Answer» Given, the first term (a) = -14, Filth term (a₅) = 2, Sum of terms (S_n) = 40 of the A.P. If the common difference is taken as d. Then, a₅ = a+ 4d ⟹ 2 = -14 + 4d ⟹ 2 + 14 = 4d ⟹ 4d = 16 ⟹ d = 4 Next, we know that S_n= \(\frac{n}{2}\)[2a + (n − 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms Now, on substituting the values in S_n ⟹ 40 = \(\frac{n}{2}\)[2(−14) + (n − 1)(4)] ⟹ 40 = \(\frac{n}{2}\)[−28 + (4n − 4)] ⟹ 40 = \(\frac{n}{2}\)[−32 + 4n] ⟹ 40(2) = – 32n + 4n² So, we get the following quadratic equation, 4n² – 32n – 80 = 0 ⟹ n² – 8n + 20 = 0 On solving by factorization method, we get 4n² – 10n + 2n + 20 = 0 ⟹ n(n – 10) + 2( n – 10 ) = 0 ⟹ (n + 2)(n – 10) = 0 Either, n + 2 = 0 ⟹ n = -2 Or, n – 10 = 0 ⟹ n = 10 Since the number of terms cannot be negative. Therefore, the number of terms (n) is 10.

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?

Answer»

Given, the first term (a) = -14, Filth term (a₅) = 2, Sum of terms (S_n) = 40 of the A.P.

If the common difference is taken as d.

Then, a₅ = a+ 4d

⟹ 2 = -14 + 4d

⟹ 2 + 14 = 4d

⟹ 4d = 16

⟹ d = 4

Next, we know that S_n= \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

Now, on substituting the values in S_n

⟹ 40 = \(\frac{n}{2}\)[2(−14) + (n − 1)(4)]

⟹ 40 = \(\frac{n}{2}\)[−28 + (4n − 4)]

⟹ 40 = \(\frac{n}{2}\)[−32 + 4n]

⟹ 40(2) = – 32n + 4n²

So, we get the following quadratic equation,

4n² – 32n – 80 = 0

⟹ n² – 8n + 20 = 0

On solving by factorization method, we get

4n² – 10n + 2n + 20 = 0

⟹ n(n – 10) + 2( n – 10 ) = 0

⟹ (n + 2)(n – 10) = 0

Either, n + 2 = 0

⟹ n = -2

Or, n – 10 = 0

⟹ n = 10

Since the number of terms cannot be negative.

Therefore, the number of terms (n) is 10.