The successive terms of an A-P- are a 1- a 2- a 3- - - Ifa 6- a 9- a 12- a 15-20- then - r -120 ar -A- 75B- 100C- 120D- 150

The correct option is B 100Let d be the common difference ⇒ a_6 + a_9 + a_12 + a_15 = 20 ⇒ a_1 + 5d + a_1 + 8d + 11d + a_1 + 14d = 20 ⇒ 4a_1 + 38d = 20 ⇒ 2a_1 ...

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Standard XII

Mathematics

Arithmetic Mean

The successiv...

Question

The successive terms of an A.P. are $a_{1}, a_{2}, a_{3}, \dots \dots . I f a_{6} + a_{9} + a_{12} + a_{15} = 20, t h e n \sum_{r = 1}^{20} a r =$

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100

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120

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150

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Solution

The correct option is B 100
Let d be the common difference $\Rightarrow a_{6} + a_{9} + a_{12} + a_{15} = 20 \Rightarrow a_{1} + 5 d + a_{1} + 8 d + 11 d + a_{1} + 14 d = 20 \Rightarrow 4 a_{1} + 38 d = 20 \Rightarrow 2 a_{1} + 19 d = 10 \Rightarrow S_{20} = [2 a + 19 d] = 10 \times 10 = 100$

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The successive terms of an A.P. are a1,a2,a3,…….Ifa6+a9+a12+a15=20, then ∑20r=1ar=

The correct option is B 100Let d be the common difference ⇒a6+a9+a12+a15=20⇒a1+5d+a1+8d+11d+a1+14d=20⇒4a1+38d=20⇒2a1+19d=10⇒S20=[2a+19d]=10×10=100

The successive terms of an A.P. are $a_{1}, a_{2}, a_{3}, \dots \dots . I f a_{6} + a_{9} + a_{12} + a_{15} = 20, t h e n \sum_{r = 1}^{20} a r =$

The correct option is B 100
Let d be the common difference $\Rightarrow a_{6} + a_{9} + a_{12} + a_{15} = 20 \Rightarrow a_{1} + 5 d + a_{1} + 8 d + 11 d + a_{1} + 14 d = 20 \Rightarrow 4 a_{1} + 38 d = 20 \Rightarrow 2 a_{1} + 19 d = 10 \Rightarrow S_{20} = [2 a + 19 d] = 10 \times 10 = 100$