Question

find sum of first four consecutive multiples of 7​

Answer 1

Answer:

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Answer 2

Answer:

$\huge \mathbb{METHOD \: \: 1} :$

AP : 7, 14 ,21 ,28

$\longmapsto \: \large \tt{a = 7 , \: d = 7 \: , n = 4 }$

Sum of n terms of an AP

$\: \: \: \: \: = \dfrac{n}{2} (2a + ( n- 1)d)$

Sum of 4 terms =

$\: \: \: \: = \dfrac{4}{2} (2 \times 7 + 3 \times 7)$

$\: \: \: \: = 2(14 + 21)$

$\: \: \: \: = 2(35) = 70$

Therefore sum of 4 consecutive multiples of 7 is 70 .

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$\huge \mathbb{METHOD \: 2} :$

First 4 multiples of 7 are 7 ,14 ,21 ,28

Sum of 4 multiples = 7 + 14 + 21 + 28

= 70

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