Question 1191995
a) The first term of an arithmetic progression is 7 and 
the common difference is 2. Find the 15th term...<pre> 
The "hard" way is to write them out starting with 7 and 
adding 2 each time
until you have 15 terms:

7,9,11,13,15,17,19,21,23,25,27,29,31,33,35.

The "easy" way is to substitute in the formula {{{a[n]=a[1]+(n-1)d}}}
where a<sub>1</sub> = first term = 7 and 
d = common difference = 2, and n = number of term = 15

{{{a[n]=a[1]+(n-1)d}}}
{{{a[15]=7+(15-1)2}}}
{{{a[15]=7+(14)2}}}
{{{a[15]=7+28}}}
{{{a[15]=35}}}
</pre>...and the sum of the first 15th term.<pre>

The "hard" way is to add them up:

7+9+11+13+15+17+19+21+23+25+27+29+31+33+35 - 315.

The "easy" way is to substitute in the formula {{{S[n]=expr(n/2)(a[1]+a[n])}}}
where a<sub>1</sub> = first term = 7 and a<sub>n</sub> = a<sub>15</sub> = 35.

{{{S[n]=expr(n/2)(a[1]+a[n])}}}
{{{S[15]=expr(15/2)(7+35)}}}
{{{S[15]=expr(15/2)(42)}}}
{{{S[15]=315)}}}

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There are two formulas for the sum of an arithmetic progression.

If you know the last term, it's easier to use the one we just used,

{{{S[n]=expr(n/2)(a[1]+a[n])}}}

But if you don't know the last term, then use the other one:

{{{S[n]=expr(n/2)(2a[1]+(n-1)d)}}}

------------------------------------------------</pre>
(b) how many terms of the sequence -9,-6,-3 must be taken that 
the sum maybe 66.<pre>

We don't know the last term, so we use the second formula, with
a<sub>1</sub>=-9 and common difference = 
(2nd term) - (1st term) =  (-6) - (-9) = -6+9 = 3

{{{S[n]=expr(n/2)(2a[1]+(n-1)d)}}}

We substitute S<sub>n</sub> = 66 and solve for n

{{{66=expr(n/2)(2(-9)+(n-1)(3))}}}
{{{66=expr(n/2)(-18+3n-3)}}}
{{{66=expr(n/2)(-21+3n)}}}
Multiply both sides by 2
{{{132=n(-21+3n)}}}
{{{132=-21n+3n^2)}}}
{{{-3n^2+21n+132=0}}}
Divide through by -3
{{{n^2-7n-44=0}}}
Factor:
{{{(n+4)(n-11)=0}}}
n+4=0;  n-11=0
  n=-4;    n=11

n can't be negative, so n = 11. It takes 11 terms to have sum 66.

Edwin</pre>