In a finite sequence S, each term after first term is 2 more than its previous term. What is the first term of the sequence S?
It is given that if the first term is x, then second term is x+2, third term is x+4 and so on.
Also, the sequence forms an ascending A. P.
1) The last term of the sequence S is 19
It is not given how many terms are there in the sequence.
The sequence can be 15, 17, 19 with first term 15 or it can be 13, 15, 17, 19 with first term 13
Insufficient.
2) The Standard deviation of all the terms of the sequence is root 40.
√40 = 2√10
Since, the sequence forms an A. P. with common difference 2, the number of terms can be determined from the Standard deviation (SD)
but the first term can not be determined.
Insufficient.
Combining 1) and 2)
For an ascending A. P. with common differnce(d) = 2 (from question stem), last term(l) = 19 (from statement 1), and known number of terms(n) (from statement 2), the first term (a) can be determined using
l = a + (n-1)d
Hence sufficent.
Chocie C is the answer.
Additional info:For an A. P. with common difference 1
S. D. = \(\sqrt{\frac{n^2-1}{12}}\) :n = number of terms of A. P.
e. g. for an A. P. with 11 terms starting from 9
i. e. 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19
n=11
Put n in the above formula
SD 1 = \(\sqrt{\frac{n^2-1}{12}}\)
= \(\sqrt{\frac{11^2-1}{12}}\)
= \(\sqrt{\frac{120}{12}}\)
= \(\sqrt{10}\)
Therefore, for any A. P. with common differnce 1 and 11 terms, the SD will be √10
For an A. P. with 11 terms and common difference 2
i. e. -1, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
the S.D. will be 2 times that of the A. P. with common difference 1
i . e. SD 2 = SD 1
SD 2 = 2 * SD 1
= 2 * √10
= √40
which is the same as given in the question.