Math, asked by InnocentBOy143, 11 months ago

Find the sum of series 72+70+68+.....+40.

Answers

Answered by Anonymous
36

SOLUTION

Given,

a=72

d= -2

l= 40

Therefore,

l= a+(n-1)d

=) 40= 72(n-1)(-2)

=) 40= 72-2n+2

=) 2n= 34

=) n= 34/2

=) n= 17

Now,

the required sum of n terms Sn

 =  >  \frac{n}{2} (a + l) \\  \\  =  >  \frac{17}{2} (72 + 40) \\  \\  =  >  \frac{17}{2}  \times 112 \\  \\  =  > 17 \times 56 = 952

hope it helps ☺️

Answered by jinadevkv
0

Answer:

952 is your answer

Step-by-step explanation:

first term, a1 = 72

common difference, d = 70 - 72 = -2

Number of terms, n = (an - a1)/d + 1 = (40-72)/-2 + 1 = 32/2 + 1 = 17

Sum = n/2 * (a1 + an) = 17/2*(72+40) = 952

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