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Insert seven numbers between $2$  and $34$, such that the resulting sequence including $2$ and $34$  is an arithmetic progression. The sum of these inserted seven numbers is  ______.

  1. $120$
  2. $124$
  3. $126$
  4. $130$
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Migrated from GO Civil 3 years ago by Arjun

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a = 2 and l = 34

A total of seven numbers are inserted between a and l so the total number of terms in A.P. will become 9.

Sum of nine numbers = n/2(a+l) =9/2(2+34) =162

Sum of inserted numbers = sum of nine numbers – (a+l) = 162 – 36 = 126

Therefore the correct answer is (C)

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The shortcut method for above questions is 

S= Sum

N = No. of terms

a = First term

l = Last term

S = (N*(a+l))/2

S = (7*(2+34))/2

S = (7*(36))/2

S = 7*18

S = 126

Hence the answer is Option C. 126

 

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Total number of terms including 2,34 = 9

WKT n/2(first term+last term)=> 9/2(2+34)=> 162

This 162 sum contains 2,34 also.

They asked for only sum of seven inserted numbers

So: 162-(2+34)=126

Option C is answer
Answer:

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