Question 1163799
<br>
If the reference you have is like most, it tells you the sum of n terms of an arithmetic sequence is<br>
{{{(n/2)(a(1)+a(n))}}}<br>
where n is the number of terms and a(1) and a(n) are the first and last terms.<br>
While that formula is valid, it is not intuitively obvious.  Why should you divide the number of terms by 2 and then multiply by the sum of the first and last terms?<br>
I VERY MUCH prefer this alternative form of the formula:<br>
{{{n((a(1)+a(2))/2)}}}<br>
In that form, the formula says the sum is the average of the terms, multiplied by the number of terms.  That makes perfect sense -- that is what average means.<br>
So let's use that formula with the two pieces of given information: the sum of the first 40 terms is 430, and the sum of the first 60 terms is 945.<br>
Use a as the first term and d as the common difference.<br>
Sum of first 40 terms is 430....<br>
number of terms: n = 40
first term: a
40th term: a+39d
sum: {{{40((a+a+39d)/2) = 430}}}  [1]<br>
Sum of first 60 terms is 945....<br>
number of terms: n = 60
first term: a
60th term: a+59d
sum: {{{60((a+a+59d)/2) = 945}}}  [2]<br>
Now solve [1] and [2] simultaneously.<br>
{{{40((a+a+39d)/2) = 430}}}
{{{2a+39d = 430/20 = 43/2}}}  [3]<br>
{{{60((a+a+59d)/2) = 945}}}
{{{2a+59d = 945/30 = 63/2}}}  [4]<br>
Comparing [3] and [4] gives us<br>
{{{20d = 20/2 = 10}}}
{{{d = 1/2}}}<br>
Substituting in [3]...<br>
{{{2a+39(1/2) = 43/2}}}
{{{2a = 4/2 = 2}}}
{{{a = 1}}}<br>
The first term of the sequence is a=1; the common difference is d=1/2.<br>
The formula for the n-th term of the sequence is<br>
{{{a(n) = a(1)+(n-1)d}}}
{{{a(n) = 1 + (n-1)(1/2) = 1 + (n-1)/2}}}<br>
CHECK:
40 terms....
1st term: 1
40th term: 1+39/2 = 41/2
sum: {{{40((1+41/2)/2) = 20(43/2) = 10(43) = 430}}} correct<br>
60 terms....
1st term: 1
60th term: 1+59/2 = 61/2
sum: {{{60((1+61/2)/2) = 30(63/2) = 15(63) = 945}}} correct<br>