# Find 30th pentagonal which is = to S30 for the arithetic serie 1+4+7+10....Please explain process

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Find the 30th term of the arithmetic series1+4+7+10+...

Note that the underlying arithmetic sequence has first term `a_1=1` and common difference `d=3` . The formula for the nth term of an arithmetic sequence is `a_n=a_1+(n-1)d` ((You are starting with the first term and then adding some number of d's; you add one less than the term number as you did not add a d to the first term))

The 30th term is `a_30=1+(30-1)3=1+87=88`

The nth term is `1+(n-1)3=1+3n-3=3n-2`

The series can be represented as `sum_(i=1)^30 3n-2`

For a small n you could add the terms:

1+4+7+10+13+16+19+22+25+28+31+34+37+40+43+46+49+52

+55+58+61+64+67+70+73+76+79+82+85+88 or you could use a graphing utility or spreadsheet to add these.

Algebraically we use the summation formulas:

`sum_(i=1)^n c=cn` for a constant c and `sum_(i=1)^n cn=c[(n(n+1)/2)]`

`sum_(i=1)^30 3n-2=3sum_(i=1)^30n-sum_(i=1)^30 2`

`=3[(30*31)/2]-2(30)`

`=45*31-60=1335`

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**The 30th number in the series is`sum_(i=1)^30 3n-2=1335` **

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