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In an AP .Given an = 4, d = 2, Sn = − 14, find n and a. - Mathematics

 In an AP .Given an = 4, d = 2, Sn = − 14, find n and a.

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Solution

Given that, an = 4, d = 2, Sn = −14

an = a + (n − 1)d

4 = a + (− 1)2

4 = a + 2n − 2

a + 2n = 6

= 6 − 2n (i)

`S_n = n/2[a+a_n]`

`-14=n/2[a+4]`

−28 = (a + 4)

−28 = (6 − 2n + 4) {From equation (i)}

−28 = (− 2n + 10)

−28 = − 2n2 + 10n

2n2 − 10n − 28 = 0

n2 − 5−14 = 0

n2 − 7n + 2n − 14 = 0

(n − 7) + 2(n − 7) = 0

(n − 7) (n + 2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we obtain

a = 6 − 2n

a = 6 − 2(7)

= 6 − 14

= −8

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APPEARS IN

NCERT Class 10 Maths
Chapter 5 Arithmetic Progressions
Exercise 5.3 | Q 3.08 | Page 112
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