Question

find the sum of series 1.3 + 1.1 + 99 + ........ + 49

Answer 1

Hey,

It's an A.P. 

where,

A= 103, D= -2

An=A+(n-1)D

49=103+(n-1)(-2)

-54=(n-1)(-2)

(n-1)=27

n=28

Now,

Sn=n/2[2A+(n-1)D]

S28 = 28/2[2(103)+(28-1)(-2)]

S28 = 14[206-54]

S28 = 14*152

S28 = 2128...ANSWER.

HOPE IT HELPS YOU:-))

Answer 2

Answer:

Correction in the question, the correct series is : 103, 101, 99, .... 49

Hence we must find the sum of the above series.

The given series is an AP.

Hence we get,

a = 103, d = -2, $a_n$ = 49, n = ?

=> 49 = 103 + ( n - 1 ) -2

=> 49 - 103 = ( n - 1 ) -2

=> -54 = ( n - 1 ) -2

=> -54 / -2 = ( n - 1 )

=> 27 = n - 1

=> n = 27 + 1 = 28 terms

Sum of AP = $\dfrac{n}{2} [ a + a_n ]$

=> Sum = $\dfrac{28}{2} [ 103 + 49 ]$

=> Sum = $\dfrac{28 }{2} [ 152 ]$

=> Sum = 28 ( 76 )

=> Sum = 2128

Hence the sum of the given series is 2128.