Question

if in any decreasing arithmetic progression sum of all its terms except the first term is equal to -36, the sum of all its terms except for the last term is 0 and the difference of the tenth and the sixth term is equal to -16,then find the first term of the series

possible answers ; 15, or 14, or 16, or 17

Answer 1

Answer:

this is the appropriate answer

Answer 2

if in any decreasing arithmetic progression sum of all its terms except the first term is equal to -36, the sum of all its terms except for the last term is 0 and the difference of the tenth and the sixth term is equal to -16,then the first term of the series is 16.

Explanation:

Sum of 'n' terms of an AP is given by

$S_{n} = \frac{n}{2} [2a+(n-1)d]$

By the given data we have,

$\frac{n}{2} [a+a_{n}] - a = -36$ ...........( 1 )

$\frac{n}{2}[a+ a_{n}] -a_{n} = 0$ ........ ( 2 )

( 1 ) - ( 2 )

$-a+a_{n} = -36\\ -a + a + (n-1)d =-36\\(n-1)d =-36$ ...... ( 3 )

Next, $a_{10} - a_{6} = -16\\a+9d-a-5d =-16\\4d = -16\\d= -4$

From equation ( 3 ) we get,

$(n-1) (-4) =-36\\(n-1) = \frac{-36}{-4}\\(n-1) = 9\\n = 9+1 = 10$

Now from equation ( 1 ) we have,

$\frac{10}{2} [a+a+(10-1)(-4)] -a = -36\\5 [2a-36 ] -a = -36\\9a -180 = -36\\9a = 180-36a = \frac{180-36}{9}$

∴ $a = 16$

Hence the first term of the series is 16.