Question

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

Answer 1

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 ☺️

Answer 2

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