Question

For an AP, if a↓n=4,d=2 and s↓n=-14 find n and a?​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \binom{a}{ \: \: \: n} = 4$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: d \: = 2$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \binom{s}{ \: \: \: n} = - 14$

$\: \: \: \boxed{ \: \: \: \: \: \: \: \: \binom{a}{ \: \: \: n} = a + (n - 1)d}$

$\: \: : ⟹ \ \: 4 = a + (n - 1)2$

$\: \: \: \: \: : ⟹4 = a + 2n - 2$

$\: \: \: \: \: \: \: \: : ⟹a + 2n = 6 \: \: \: \: \: \: \: \: \: \: (i)$

$\: \: \: \boxed{ \binom{s}{ \: \: \: n} = \frac{n}{2} (2a + (n - 1)d) }$

$\: \: : ⟹ \: - 14 = \frac{n}{2} (2a + (n - 1)2)$

$\: \: \: \: \: :⟹ - 28 = n(2a + 2n - 2)$

$\: \: \: \: \: \: \: \: : ⟹ \frac{ - 28}{ \: \: n} = 2a + 2n - 2 \: (ii)$

$\: \: \: \: \: \: \: \: \: \: \: from \: equation \: 1$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed {a = 6 - 2n}$

$\: \: substituting \: value \: of \: a \: in \: eq {}^{2}$

$: ⟹ \: \: \frac{ - 28}{ \: \: n} = 2(6 - 2n) + 2n - 2$

$: ⟹ \frac{ - 28}{ \: \: n} = 12 - 4n + 2n - 2$

$\: \: \: \: : ⟹ \frac{ - 28}{ \: \: n} = 10 - 2n$

$\: \: \: \: \: \: \: \: :⟹ \: 10n - 2 {n}^{2} = - 28$

$\: \: \: \: : ⟹2 {n}^{2} - 10n - 28 = 0$

$\: \: \: \ : ⟹2 {n}^{2} + 4n - 14n - 28 = 0$

$\: : ⟹ \: 2n(n + 2) - 14(n + 2) = 0$

$\: \: \: \: \: \ : ⟹ \: \: (2n - 14)(n + 2) = 0$

$\: \: \: \: \: \: \: : ⟹n = 7 \: and \: - 2$

$as \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: a = 6 - 2n$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: n = 7 \\ \: \: \: \: \: \: \: \: \: \: \: \ :⟹ \: \: \: \: \: a = 6 - 14 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: :⟹ \: \: \: \boxed{a = - 8 }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: n = - 2 \\ \: \ \: \: \: \: \: \: \: \: \: : ⟹ \: a = 6 + 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: : ⟹ \boxed{a = 8}$

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Answer 2

Given:-

• $\sf{a_n = 4}$
• d = 2
• $\sf{S_n = -14}$

To Find:-

The value of n and a

Solution:-

We know,

$\sf{a_n = a + (n-1)d}$

Substituting the value,

$\sf{4 = a + (n-1)\times 2}$

= $\sf{4 = a + 2n - 2}$

= $\sf{4+2 = a+2n}$

= $\sf{6 = a+2n}$

=> $\sf{a = 6-2n \longrightarrow(i)}$

Now,

Now,We also know,

$\sf{S_n = \dfrac{n}{2}[2a + (n-1)d]}$

= $\sf{-14 = \dfrac{n}{2}[2a + (n-1)\times 2]}$

= $\sf{-14\times 2 = n[2(6-2n) + 2n - 2]\:\:\:\:\:[From\:Eq.(i)]}$

= $\sf{-28 = n[12- 4n + 2n - 2]}$

= $\sf{-28 = n[12 - 2n - 2]}$

= $\sf{-28 = n[10 - 2n]}$

= $\sf{-28 = 10n - 2n^2}$

= $\sf{-28 = 2(5n - n^2)}$

= $\sf{\dfrac{-28}{2} = 5n - n^2}$

= $\sf{-14 = 5n - n^2}$

= $\sf{n^2 - 5n - 14 = 0}$

By splitting the middle term,

= $\sf{n^2 - 7n + 2n - 14 = 0}$

= $\sf{n(n-7) + 2(n-7) = 0}$

= $\sf{(n-7)(n+2) = 0}$

Either,

$\sf{n-7 = 0}$

= $\sf{n = 7}$

Or,

$\sf{n+2 = 0}$

= $\sf{n= -2}$

Negative value of n cannot be taken.

$\sf{\therefore}$ n = 7

Putting the value of n in Eq.(i)

= $\sf{a=6-2n}$

= $\sf{a=6-2\times 7}$

= $\sf{a=6-14}$

= $\sf{a= -8}$

Therefore,

The value of a = -8

The value of n = 7

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