Question

28: nth of an A.P. is 4 and common difference is
2 at the sum of n terms is (-14) then (A)
First
term of the A.P​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Question :-

nth of an A.P. is 4 and common difference is

2 at the sum of n terms is (-14) then (A)

First term of the A.P.

Answer :-

• a = 10 or
• a = -8

Step by step explanation :-

We go step by step,

Given that,

nth of an A.P. is 4.

Hence,

$\rm:\longmapsto{}a + (n - 1)d = 4$

Also we have,

=> difference = 2

Now,

$\rm:\longmapsto{a + (n - 1)2 = 4} \\ \\\rm:\longmapsto{a + 2n - 2 = 4} \: \: \: \\ \\ \rm:\longmapsto{a + 2n = 4 + 2} \: \: \\ \\ \rm:\longmapsto \red{a = 6 - 2n} \: \: \: \: \: \: \: \: \:$

Also we have given that,

The sum of n terms is (-14).

Then,

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

Substitute the value of d and a,

$\rm:\longmapsto{ - 14 = \frac{n}{2} \{2(6 - 2n) + (n - 1)2 \} } \\ \\\rm:\longmapsto{ - 14 = \frac{n}{2} (12 - 4n + 2n - 2 ) } \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{ - 14 = \frac{n}{2}(10 - 2n) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{ - 14 = \frac{n(10 - 2n)}{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{ - 14 = \frac{10n - 2n {}^{2} }{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{ - 14 = \frac{ \cancel{2}(5n - n {}^{2}) }{ \cancel2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{ - 14 = 5n - n {}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto \red{n {}^{2} - 5n - 14 = 0} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now, we factorise the equation,

$\rm:\longmapsto{n {}^{2} - 5n - 14 = 0} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{n {}^{2} - 7n + 2n - 14 = 0 } \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{n(n - 7) + 2(n - 7) = 0} \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{(n - 7)(n + 2) = 0} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{n - 7 = 0 \: \: \: \: \: or \: \: \: \: \: \: n + 2 = 0} \\ \\ \rm:\longmapsto \red{n = 7 \: \: \: \: \: or \: \: \: \: \: n = - 2} \: \: \: \: \: \: \: \: \: \: \:$

Now, we have,

=> n = 7

=> n = -2

Substituting n = 7 in a = 6 - 2n,

$\rm:\longmapsto{a = 6 - 2n} \: \: \: \\ \\ \rm:\longmapsto{a = 6 - 2(7)} \\ \\ \rm:\longmapsto{a = 6 - 14} \: \: \: \\ \\ \rm:\longmapsto \red{a = - 8} \: \: \: \: \: \: \: \: \:$

And substituting n = -2 in a = 6 - 2n,

$\rm:\longmapsto{a = 6 - 2n} \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{a = 6 - 2( - 2)} \: \\ \\ \rm:\longmapsto{a = 6 + 4} \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto \red{a = 10} \: \: \: \: \: \: \: \: \: \: \: \: \:$

The first term (a) is :

=> a = 10

or

=> a = -8

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