The sum of the first six terms of an arithmetic sequence is 93 and the sum of the first two terms is 430. Find: ?

a. the sum of the first n terms

b. the smallest value of n for the sum to exceed 1000

1 Answer
May 7, 2018

See below.

Explanation:

A generic arithmetic sequence is formed by terms like #an+b#, with #n \in \mathbb{N}#. The sum of the first six terms can be written as

#sum_{i=1}^6 (ai+b) = asum_{i=1}^6 i+sum_{i=1}^6b = asum_{i=1}^6 i+6b#

Now, the sum of the first #n# numbers is #\frac{n(n+1)}{2}#, so you have

#sum_{i=1}^6 i = \frac{6\cdot 7}{2} = 21#

The first equation of the system is thus

#21a+6b=93#

We also know about the sum of the first two terms, namely

#(a+b)+(2a+b) = 3a+2b = 430#

Put the two together, and you have

#a = -399/4, b = 2917/8#

NOTE: are you sure this is not a typo? I tried to solve the same equation, but with #43# instead of #430#, and got a much simpler solution:

#a = -3, b = 26#

So I'm assuming this is the correct one.

Now, the sum of the first #n# terms is just

#sum_{i=1}^n (-3i+26) = -3sum_{i=1}^n i+sum_{i=1}^n26 = -3\frac{n(n+1)}{2} + 26n#

Also, since #a# is negative, the terms of the sum are positive only if #1\le a \le 8#, and the series doesn't reach #1000#