Question 1010798
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If the 4th term of an arithmetic series is 62 and the 14th term is 122, determine the sum of the first 30 terms.
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From the condition, we have two equations for 4-th and 14-terms of the AM:

{{{a[4]}}}  =  62   and
{{{a[14]}}} = 122.

Or

{{{a[1] + 3*d}}}  =  62,   (1)
{{{a[1] + 13*d}}} = 122.   (2)

Distract the equation (1) from the equation (2). You will get

10*d = 122 - 62 = 60.

Hence, d = 6. Thus the common difference of the given AM is 6.

Having this, you can easily find the first term of the AM from (1). It is

{{{a[1]}}} = 62 - 3*6 = 44.

Now, when you know everything about the given AM, you can easily calculate the sum of the first n terms. 
Use the formula for the sum of the first n terms of arithmetic progression
(see the lesson <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Arithmetic-progressions.lesson>Arithmetic progressions</A> in this site). The sum is 

{{{S[30]}}} = {{{( a[1] + ((n-1)*d)/2 )*n }}} = {{{( 44 + ((30-1)*6)/2 )*30 }}} = 3930.
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