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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 2 years ago by Arjun

7 Answers

Best answer
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We can insert the seven numbers in such a way that, so we get the arithmetic progression including $2$ and $34.$

 We get the arithmetic progression $2,6,10,14,18,22,26,30,34$

The inserted seven numbers are $6,10,14,18,22,26,30$

Here, $a = 6,d = 10-6 = 4,l = 30,n = 7$

$\therefore$ The sum of the arithmetic progression $S_{7} = \frac{7}{2}\left(6 + 30 \right) = 7 \cdot 18 = 126.$

So, the correct answer is $(C).$
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$\therefore$ Correct option is $\large C$

5 votes
5 votes

Answer: Option C. 126

Edit: {162 – (2+34)} = 126

2 votes
2 votes
the first term of the sequence will be 2

the last term of the sequence will be 34

here it is asked that we insert 7 numbers in between 2 _ _ _ _ _ _ _ 34

so now the total no. of elements in sequence became 9

so according to the AP sum formula  Sum  =  N/2 (A+L)

here A = 2, L=34 and N = 9;

so sum of 7 inserted numbers = N/2(A+L) – (A+L)  =  9/2(2+34) – (2+34) = 126
 

this is I guess a very straightforward answer….

 

P.S. : This is very first time i am writing answer on GO …. so please correct me if im wrong :)
Answer:

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