"Let a1, a2, ..., an be a given A.P. whose common difference is an integer and Sn = a1 + a2 + .... + an. If a1 = 1, an = 300 and 15 $\le$ n $\le$ 50, then the ordered pair (Sn-4, an–4) is equal to:"

Question

Accepted Answer

"${a_n} = {a_1} + (n - 1)d$

$\Rightarrow 300 = 1 + (n - 1)d$

$\Rightarrow (n - 1)d = 299 = 13 \times 23$

since, n $\in$[15, 50]

$\therefore$ n = 24 and d = 13

${a_{n - 4}} = {a_{20}} = 1 + 19 \times 13 = 248$

$\Rightarrow {a_{n - 4}} = 248$

${S_{n - 4}} = {{20} \over 2}\{ 1 + 248\} = 2490$"

Let a₁, a₂, ..., an be a given A.P. whose
common difference is an integer and
S_n = a₁ + a₂ + .... + a_n. If a₁ = 1, a_n = 300 and 15 $\le$ n $\le$ 50, then
the ordered pair (S_n-4, a_n–4) is equal to:

Solution

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Let a1, a2, ..., an be a given A.P. whose common difference is an integer and Sn = a1 + a2 + .... + an. If a1 = 1, an = 300 and 15 $\le$ n $\le$ 50, then the ordered pair (Sn-4, an–4) is equal to:

Solution

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More Sequences and Series questions

Drill 25 more like these. Every day. Free.

Let a₁, a₂, ..., an be a given A.P. whose
common difference is an integer and
S_n = a₁ + a₂ + .... + a_n. If a₁ = 1, a_n = 300 and 15 $\le$ n $\le$ 50, then
the ordered pair (S_n-4, a_n–4) is equal to: