Question

What is the sum of first 10 terms of an A.P 41,36,31,26....?

Answer 1

Answer:

P -----> 41, 36, 31, 26, ...........

So, First term = a = 41

d = 36 - 41 = - 5

n = 10

We know,

Sn = a + ( n - 1 ) d

S10 = 41 + ( 10 - 1 ) * (-5)

S10 = 41 + ( - 45 )

S10 = - 4

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Answer 2

$\huge{\underline{\underline{\red{\bf{Given:}}}}}$

• A AP is given , 41,36,31,26,21,.........

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$\huge{\underline{\underline{\red{\bf{To\:Find:}}}}}$

• The sum of first 10 terms of the AP .

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• We will use formula to find the sum of n terms of AP .

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Given AP is 41,36,31,26,21,.....

Sum of this AP will be ,41+36+31+26+21.....+10terms.

Let the sum of AP be denoted by S.

Now sum of n terms of AP is given by ,

${\underline{\boxed{\red{\bf{\leadsto S_{n}=\dfrac{n}{2}[2a+(n-1)d]}}}}}$

where

• ☞ n is number of terms.
• ☞ a is the first term.
• ☞ d is the common difference.
• ☞ $\sf{S_{n}}$ is the sum of series.

Here ,

• ☞ First term is 41.
• Common Difference is (-5).
• Number of terms = 10.

⇒S = 10/2 [ 2 ×41 +(10-1)×(-5)]

⇒S = 5 [ 82 + 9× (-5)]

⇒ S = 5 [ 82 -45]

⇒S = 5 × 37

⇒S = 185

Hence the required sum is 185.