Question

If a1, a2, a2..... an are in Arithmetic Progression, whose common difference is an integer such that a1 = 1, an = 300 and n ∈ [15, 50], then (Sₙ₋ ₄, aₙ₋₄) is :

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$a_1, a_2, a_2..... a_n$

are in Arithmetic Progression

Let d be the Common Difference

So

$a_n = a_1 + (n - 1)d$

$\implies \: 300 = 1 + (n - 1)d$

$\implies \: (n - 1)d = 299$

Now the factors of 299 are 1, 13, 23, 299

So we can write

$299 = 23 \times 13$

$\because \: n \in [ 15, 50]$

and d is an integer

So From above we can say that

$n - 1 = 23 \: \: \: and \: \: d = 13$

$\implies \: n = 24 \: \: \: and \: \: d = 13$

So

$S_{n-4}$

$= S_{24-4}$

$= S_{20}$

$= \displaystyle \: \frac{20}{2} \{2 \times 1 + (20 - 1) \times 13 \}$

$= 10 \times (2 + 247)$

$= 2490$

Also

$a_{n-4}$

$= a_{24-4}$

$= a_{20}$

$=a_{1} + (20 - 1) \times 13$

$= 1 + 247$

$= 248$

So

$(S_{n-4} \: , \: a_{n-4}) = (2490 \: , \: 248)$

Answer 2

Answer:

b

Step-by-step explanation:

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