Question

if an = 4, d=2,sn=-14 find the value of n and a

Answer 1

an=a+(n-1)d=4

or, a+(n-1)2=4

or, a+2n-2=4

or, a+2n=4+2

or, a+2n=6

or, a=6-2n

=n/2[2a+(n-1)d]=-14

or, n/2[2a+(n-1)2]=-14

or, an+n(n-1)=-14

or, (6-2n)n+n²-n=-14

or, 6n-2n²+n²-n=-14

or, -n²+5n=-14

or, n²-5n=14

or, n²-5n-14=0

or, n²-7n+2n-14=0

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

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

either, n-7=0

or, n=7

or, n+2=0

or, n=-2

∵, n can not be negative ;

∴, n=7

∴, a=6-(2×7)

=6-14

=-8

∴, n=7 and a=-8 Ans.

Answer 2

GIVEN : an = 4 d = 2, Sn = - 14

FIND : Value of n and a.

SOLUTION :

We know that..

an = a + (n - 1)d

Put the known values in above formula to find the value of n.

=> 4 = a + (n - 1)2

=> 4 = a + 2n - 2

=> 4 + 2 = a + 2n

=> 6 = a + 2n

=> 6 - 2n = a _________ (eq 1)

Now..

Sn = n/2 (a + an)

Put the known values in above formula

=> - 14 = n/2 [(6 - 2n) + 4]

=> - 14 = n/2 (6 - 2n + 4)

=> - 14 = n/2 (10 - 2n)

=> -14(2) = n(10 - 2n)

=> - 28 = 10n - 2n²

=> 2n² - 10n - 28 = 0

Take 2 common from L.H.S.

=> 2(n² - 5n - 14) = 0

=> n² - 5n - 14 = 0

The above equation is in the form ax² + bx + c and is equal to 0.

i.e.

ax² + bx + c = 0

Now, we have to find a integer whose sum is "b" i.e. (-5) and product is "c" i.e. (-14)

=> n² - 5n - 14 = 0

=> n² - 7n + 2n - 14 = 0

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

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

=> n = 7, n = -2 (Neglected)

Put value of n in (eq 1)

=> 6 - 2(7) = a

=> 6 - 14 = a

=> a = - 8

Answer :-

$\bold{\underline{a\:=\:-8\:and\:n\:=\:7}}$