Quant - AP/GP

Question 1

For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by ​open paren negative 1 close paren raised to the k plus 1 power times open paren the fraction with numerator 1 and denominator 2 to the kth power close paren​ . If T is the sum of the first 10 terms in the sequence, then T is

greater than 2
between 1 and 2
between ​one half​ and 1
between ​one fourth​ and ​one half​
less than ​one fourth

Answer 1

ANS = D

First of all we see that there is set of 10 numbers and every even term is negative.

Second it’s not hard to get this numbers: 1212, −14−14, 1818, −116−116, 132132… enough for calculations, we see pattern now.

And now the main part: adding them up is quite a job, after calculations you’ll get 34110243411024. You can add them up by pairs but it’s also time consuming. Once we’ve done it we can conclude that it’s more than 1414 and less than 1212, so answer is D.

BUT there is shortcut:

Sequence 1212, −14−14, 1818, −116−116, 132132… represents geometric progression with first term 1212 and the common ratio of −12−12.

Now, the sum of infinite geometric progression with common ratio |r|<1|r|<1, is sum=b/1−r, where b is the first term.

So, if the sequence were infinite then the sum would be: 121−(−12)=13121−(−12)=13

This means that no matter how many number (terms) we have their sum will never be more then 1313 (A, B and C are out). Also this means that the sum of our sequence is very close to 1313 and for sure more than 1414 (E out). So the answer is D.

Answer: D.