Find the sum of the following APs- -i- 2- 7- 12 - to 10 terms- -ii- - 37- - 33- - 29 - to 12 terms -iii- 0-6- 1-7- 2-8 - to 100 terms -iv- 1-15- 1-12- 1-10- - - to 11 terms - Maths Q-A

Finding the sum of the given AP :Given, n=11, a=1/15 and d=a_2 a_1=1/12 1/15=1/60Sum of an AP (S_n) is given as,S_n=n/2[2a+(n 1)d]0ex0ex =11/2[2×1/15+(11 1)· ...

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Byju's Answer

Standard X

Mathematics

Sum of 'n' Terms of an Arithmetic Progression

Find the sum ...

Question

Find the sum of the following AP: $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, . . .$ to $11$ terms

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Solution

Finding the sum of the given AP :
Given, $n = 11$ , $a = \frac{1}{15}$ and $d = a_{2} - a_{1} = \frac{1}{12} - \frac{1}{15} = \frac{1}{60}$
Sum of an AP ( $S_{n}$ ) is given as,
$S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{11}{2} [2 \times \frac{1}{15} + (11 - 1) \cdot \frac{1}{60}] = \frac{11}{2} (\frac{2}{15} + \frac{10}{60}) = \frac{11}{2} \times \frac{9}{30} = \frac{33}{20}$
Therefore, the sum of the AP $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, . . .$ to $11$ terms is $\frac{33}{20}$

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Find the sum of the following AP: 115,112,110,... to 11 terms

Finding the sum of the given AP :Given, n=11, a=115 and d=a2-a1=112-115=160Sum of an AP (Sn) is given as,Sn=n22a+(n-1)d=1122×115+(11-1)·160=112215+1060=112×930=3320Therefore, the sum of the AP 115,112,110,... to 11 terms is 3320

Find the sum of the following AP: $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, . . .$ to $11$ terms