Question

If nth term is 4and difference is 2 and sum of n terms is -14 then find a

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{First\:term=-8}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Last \: term(a_{n}) = 4 \\ \\ \tt: \implies Common \: difference = 2 \\ \\ \tt: \implies Sum \: of \: nth \: term(s_{n}) = -14 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies First \: term( a_{1}) = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies a_{n} = a + (n - 1)d \\ \\ \tt: \implies 4 = a + (n - 1) \times 2 \\ \\ \tt: \implies 4 = a +2n - 2 \\ \\ \tt: \implies 6 = a + 2n \\ \\ \tt: \implies 6 - a = 2n \\ \\ \tt: \implies n = \frac{6 - a}{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{n} = \frac{n}{2} (2a + (n - 1) \times d \\ \\ \tt: \implies - 14 = \frac{6 - a}{2 \times 2} (2a + ( \frac{6 - a}{2} - 1) \times 2) \\ \\ \tt: \implies - 14 = \frac{6 - a}{4} (2a + ( \frac{6 - a - 2}{2} ) \times 2 )\\ \\ \tt: \implies - 14 = \frac{6 - a}{4} (2a + 4 - a) \\ \\ \tt: \implies 14 \times 4 = (a - 6)(a + 4) \\ \\ \tt: \implies 56 = {a}^{2} - 2a - 24 \\ \\ \tt: \implies {a}^{2} - 2a - 80 = 0 \\ \\ \tt: \implies {a}^{2} - 10a + 8a - 80 = 0 \\ \\ \tt: \implies a(a - 10) + 8(a - 10) = 0 \\ \\ \tt: \implies (a + 8)(a - 10) = 0 \\ \\ \green{\tt: \implies a = - 8 \: and \: 10}\\\\ \tt Note-Neglect\:value\:of\:a=10\:because\:it\:gives\:negative\:terms\\\\ \green{\tt\therefore First\:term\:is\:-8}$

Answer 2

$\tt{\huge{\purple{ ---------------- }}}$

QUESTION :

If nth term is 4and difference is 2 and sum of n terms is -14 then find a.

SOLUTION :

We know the following Formulae...

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

=> 4 = a + 2 { n - 1 }

=> a + 2n = 6

=> n = { 6 - a } / { 2 }

Now, we have that the sum of n terms is - 14.

We know that :

S_{n} = { n}/2 [ 2 a + ( n - 1 ) d ]

=> -14 = { 6 - a } / 4 [ 2a + { 4 - a } / 2 × 2 ]

=> -14 = { 6 - a } / 4 [ a + 4 ]

=> 56 = { a - 6 } { a + 4 }

=> a ^ 2 - 2a - 80 = 0

=> a ^ 2 - 10 a + 8 a - 80 = 0

=> a ( a - 10 ) + 8 ( a - 10 ) = 0

=> ( a + 8 )( a - 10 ) = 0

Now, we need to find the Zeroes.

( a + 8 ) = 0

=> a = -8

( a - 10 ) = 0

=> a = 10

So we obtain two values of A, -8 and 10 .

ANSWER :

The two required values of A are -8 and 10.