Q. 3 F4.0( 22 Votes )

# In an AP :

given a_{n} = 4, d = 2, s_{n} = -14. Find n and a

Answer :

Given, series in A.P.

n^{th} term in the AP a_{n} = 4,

Common difference (d) = 2

Sum of the series is s_{n} = -14

a_{n} = a +(n-1)d

4 = a + (n-1)2

6 = a + 2n

a = 6 – 2n ⇒ __1__

Sum of the series (s_{n}) = [2a + (n-1) d]

-14 = [ 2×(6-2n) + (n-1)2]

-14 = n[5-n]

n^{2} -5n -14 = 0

n^{2} -7n + 2n -14 = 0

n(n-7) + 2(n-7) = 0

(n-7)(n+2) = 0

n-7 = 0 ⇒ n = 7 (or) n+2 = 0 ⇒ n = -2

since n should be a positive integer

we take n = 7

substitute in equation __1__

a = 6 – 2n

a = 6 – 2(7)

a = -8

∴ first term is -8

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Show that a_{1}, a_{2},…..a_{n} form an AP where is defined as below :

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