Question

Find the sum of the series 41+36+31+... to 15 terms

Answer 1

The Sum is 90.



41+36+31+... is an Arithmetic Progression (A.P.)


Here, 

First Term = a = 41
Common Difference = d =  - 5

We have to find the sum of first 15 terms. 
So, n = 15

Sum of n terms of AP is given by:

$\boxed{S_n = \frac{n}{2}(2a+(n-1)d)} \\ \\ \\ \implies S_{15} = \frac{15}{2}(2(41)+(15-1)\times(-5)) \\ \\ \\ \implies S = \frac{15}{2}(82 - 5\times 14) \\ \\ \\ \implies S = \frac{15}{2}(82-70) \\ \\ \\ \implies S = \frac{15}{2} \times 12 \\ \\ \\ \implies S = 15 \times 6 \\ \\ \\ \implies \boxed{S=90}$

Answer 2

As we know. that

"41+36+31+.......to 15 terms. is a ARITHMETIC
PROGRESSION (AP)

so now,

Here,

_________

● First term = a = 41
_________

__________

● Common difference = d
=> first term - second term
=> 36 - 41 = -5
__________

and also given

● Total term = n = 15 terms
__________-

Now,

$Sum \: of \: the \: n \: terms \: in \: AP \: = S _{n} \\ \\ = > S _{n} = \frac{n}{2} [2a + (n - 1)d] \: \: \: \: \\ \\ = > S _{15} = \frac{15}{2} [2 \times 41 \: + (15 - 1)( - 5)] \\ \\ = > S _{15} = \frac{15}{2} (82 \: - 70) \\ \\ = > S _{15} \: = \frac{15}{2} \times 12 \\ \\ = > S _{15} = 15 \times 6 \\ \\ = > S _{15} \: = 90 \: \: \: \: \: \: \: \: .............Answer$

HENCE, SUM OF THE 15th terms of given AP = 90

DEVIL_KING ▄︻̷̿┻̿═━一