How many terms of the arithmetic sequence #{1,3,5,7,...}# will give a sum of #961#?

1 Answer
Dec 7, 2015

Answer:

#31# terms

Explanation:

For an arithmetic sequence with initial value #a# and a difference between terms of #d#
the sum of the first #n# terms is given by the formula
#color(white)("XXX")Sigma = n/2*(2color(cyan)(a)+(n-1)color(green)(d))#

For the given sequence
#color(white)("XXX")color(cyan)(a=1)#
and
#color(white)("XXX")color(green)(d=2)#

We are told that the required sum is #961#

So
#color(white)("XXX")961=n/2*(2(color(cyan)(1))+(n-1)(color(green)(2)))#

#color(white)("XXX")961=n/color(red)(cancel(color(black)(2)))*color(red)(cancel(color(black)(2))) +n/color(blue)(cancel(color(black)(2)))(n-1)(color(blue)(cancel(color(black)(2))))#

#color(white)("XXX")961 = n+n^2 - n#

#color(white)("XXX")n^2 =961#

#color(white)("XXX")n=31#