how many terms are there in the ap whose first and fifth terms are -14 and 2 and the sum of there ap is 40
Answers
Step-by-step explanation:
Given, the first term (a) = -14, Filth term (a5) = 2, Sum of terms (Sn) = 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
Sn = n 2 n2[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 Sn
⟹ 40 = n 2 n2[2(−14) + (n − 1)(4)]
⟹ 40 = n 2 n2[−28 + (4n − 4)]
⟹ 40 = n 2 n2[−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