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An arithmetic progression (AP) has 4 as its first term. What is the common difference if the sum of the first 12 terms is 2 times the sum of the first 8 terms?

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Let $d$ be the common difference. The sum of the first $12$ terms is then $$4 + (4 + d) + (4 + 2d) + \ ... \ + (4 + 11d)$$

Recall now the key fact that the sum of the first $n$ integers is $\frac{(n+1)n}{2}$. Obviously you can write the expression for the sum of the first $8$ terms similarly and it should now reduce just to arithmetic.

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So I got the solution using the above hints by pathomps and Cameron: assuming T1=a=4 d=difference Sn =sum of nterms l=last term we know 2.S8=S12 for the first 8 terms S8=n/2.(a+l) =4(4+(4+7d) = 32+28d for the first 12 terms S12 = n/2.(a+l) =6(4+(4+11d)) = 6(8+11d) = 48+66d But 2.(S8) = S12 therefore 2(32+28d) = 48+66d 64+56d=48+66d 64-48=66d-56d 10d=16 d=1.6 – Sylvester Jul 2 '13 at 7:18

Hint: Let $d$ be the common difference. The first $8$ terms are $4,4+d,4+2d,...,4+7d.$ Do you know how to find their sum? (If you aren't sure how to add up the terms with the $d$s, you might look up the triangular numbers.) The first $12$ terms are $4,4+d,4+2d,...,4+11d.$ The same approach will allow you to find that sum.

What does the relationship between the two sums need to be? Solve the resulting equation for $d$.

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If $4$ is the first term and $d$ is common difference, the sum of first $n$ terms is $\frac n2\cdot \{2\cdot4+(n-1)d\}$

So, we have $$\frac{\frac {12}2\cdot \{2\cdot4+(12-1)d\}}{\frac 82\cdot \{2\cdot4+(8-1)d\}}=\frac21$$

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Using the same principle highlighted above, I am not able to get the first term using the same principle on this case where rhe 2nd term is 11 and the 11th term is -7. I wonder whether the principle of a, a+d, a+2d works in all circumstance. – Sylvester Jul 2 '13 at 7:35
@Sylvester, That is the definition of Arithmetic Series. Here the first term $=4,$ and the common difference $=d=\frac85\implies$ the $n$ terms $=4+(n-1)\frac85$ and the sum of $n$ terms is $\frac n2\cdot \{2\cdot+(n-1)\frac85\}$ – lab bhattacharjee Jul 2 '13 at 8:10
thanks @lab-bhattacharjee – Sylvester Jul 2 '13 at 17:26

Another way of looking at this, which is less general - the sum of the first eight terms is equal to the sum of the next four (this works because the multiple in the question is $2$, which makes things easy) so:$$4+(4+d)+(4+2d)+\dots (4+7d)=(4+8d)+(4+9d)+(4+10d)+(4+11d)$$

I reckon this is simpler arithmetic - lots can be easily cancelled.

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thanks @mark-bennet – Sylvester Jul 2 '13 at 17:26

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