Question

16. The first and last terms of an A.Pare 1 and 11.if the sum of its terms is 36 , then the
number of terms is.............
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Answer 1

$a = 1 \\a_{n} = 11 \\s_{n} = 36\\ = \frac{n}{2} [ a + a_{n} ]\\ 36 = \frac{n}{2} [ 1 +11]\\ 36 = n*6\\ therefore \\ n = 6 \\\\so the no . terms ARE 6 \\\\hope it will help you \\\\thank you.$

Answer 2

Given :-

Ist term of AP = 1

Last term of AP = 11

Sum of it's terms = 36

To Find :-

The number of terms in the given AP.

Solution :-

We have,

Ist term = a = 1

last term = l = 11

Sum S = 36

${\sf{S\: =\: {\dfrac{n}{2}}\: (a\: +\: l)}}$

On inserting the values in the formula

We get ,

${\sf{36\: =\: {\dfrac{n}{2}}\: (1\: +\: 11)}}$

${\sf{\implies\: 36\: =\: {\dfrac{n}{2}}\: (1\: +\: 11)}}$

${\sf{\implies\: 36\: =\: {\dfrac{n}{2}}\: \times\: 12}}$

${\sf{\implies\: 36\: =\: 6n}}$

${\sf{\implies\: n\: =\: {\dfrac{36}{6}}}}$

${\sf{\implies\: n\: =\: 6}}$

Therefore, number of terms in the given AP is 6