Thread: The hardest Question on the Exam!

1. The hardest Question on the Exam!

There was a question that said: In an artithmetic sequence t(9) + t(22)= 61 what is the sum of the first 30 terms. Now i solved this question and i got an answer of 915. However I believe that I solved it a different way than my teacher because he said something about paring "things" to find the answer. If you know how to solve it that way please share it with me !

Either way, here is my solution:

$t(9) = 61 - t(22)$
Between t(9) and t(22) is 13 terms, so the sum of those terms would be:
$Sum13= 13/2(61-t(22) + t(22)= 396.5$

Now $t(9) +t(22) = 61$ is also the same as $t(31) = 61$

So, if t(31) = 61 , t(9) would then be = 61 - 22d and t(22) = 61 - 9d

Now sub those two new values for t9 and t22 and solve for the difference using the calculated sum 396.5!
$Sum13=13/2(61-22d+61-9d), 396.5= 6.5(122-31d), 61=122-31d, 61/31= d$

Now since d = 61/31 i can find the first term of the sequence. If t31= 61, t1=61-30d

So, t(1)= 61/31

Finally the sum of the 30 terms in this serires:

Sum30= 15(2(61/31)+(29)*61/31)
= 915

2. Re: The hardest Question on the Exam!

t9 + t22 = 61
OR [ a + 8d ] + [ a + 21d] = 61 [ Because tn = a + ( n-1 ) d where ‘a’ is the first- that is 2a + 29d = 61 ---(1) - term and ‘d’ the common difference. ]
Now sum of n terms of AP is given by Sn = n/2 [ 2a + ( n-1)d]
Thus s30 = 15 [ 2a+ 29d] = 15 x 61 Using equation (1)
Therefore Sum of the first 30 terms = 15 x 61 = 915