An algebra problem by Aman Joshi

Algebra Level 2

There are 10 numbers in the certain arithmetic progression. The sum of first 3 numbers is 321. The sum of last 3 numbers is 405. Find the sum of all the 10 terms.

1210 1256 1221 1230

Aman Joshi by Aman Joshi , 18, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Edwin Gray
Aug 10, 2018

Let a_1 = a; Then a + (a + d) + (a + 2d) = 321, or 3a + 3d = 321 and a + d = 107.(a + 7d) +(a + 8d) + (a + 9d) = 405, or 3a + 24d = 405, and a + 8d = 135. Subtracting, 7d = 28, or d = 4, and from the equation a + d = 107, we have a = 103. Then S = (10/2)(103 + 139) = 1210. Ed Gray

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Kushal Dey
Nov 3, 2017

If a1+a2+a3=231, then a2=107 and a8+a9+a10=405,then a8=135. Hence we have values of a2 and a9. From here we can say d=4. Hence a1=a2-d=107-4=103, then s10=1210

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I think you meant a9=135

Richard Costen - 3 years, 7 months ago

Log in to reply

Sorry, that was a typing mistake.

Kushal Dey - 3 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...