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:

Sequences and Series PYQ — JEE Main 2020 · 4 Sep · Shift 2 · Q36 with Solution

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Correct answer: C

For an arithmetic progression, $a_n=a_1+(n-1)d$ where $d$ is the common difference.

Given:

\quad a_n=300$$ So, $$300=1+(n-1)d$$ $$299=(n-1)d$$ Since $d$ is an integer, $(n-1)$ must be a divisor of $299$. 2. **Factorize $299$** $$299=13\times 23$$ So the positive divisors of $299$ are: $$1,\ 13,\ 23,\ 299$$ Thus, $$n-1\in\{1,13,23,299\}$$ which gives $$n\in\{2,14,24,300\}$$ Given $15\le n\le 50$, the only possible value is $$n=24$$ 3. **Find the common difference** Using $$299=(n-1)d$$ $$299=23d$$ $$d=13$$ 4. **Find $S_n$** Sum of first $n$ terms of an A.P. is $$S_n=\frac{n}{2}(a_1+a_n)$$ So, $$S_{24}=\frac{24}{2}(1+300)=12\cdot 301=3612$$ 5. **Interpret the required pair** The pair asked is $(S_{n-4}, a_{n-4})$. Since $n=24$, we need $$(S_{20}, a_{20})$$ 6. **Find $a_{20}$** $$a_{20}=a_1+19d=1+19\cdot 13=1+247=248$$ 7. **Find $S_{20}$** $$S_{20}=\frac{20}{2}(a_1+a_{20})=10(1+248)=10\cdot 249=2490$$ So, $$(S_{n-4}, a_{n-4})=(2490,248)$$ 8. **Match with the options** This is **Option C**.

View written solutionFree

Correct answer: C

For an arithmetic progression, $a_n=a_1+(n-1)d$ where $d$ is the common difference.

Given:

\quad a_n=300$$ So, $$300=1+(n-1)d$$ $$299=(n-1)d$$ Since $d$ is an integer, $(n-1)$ must be a divisor of $299$. 2. **Factorize $299$** $$299=13\times 23$$ So the positive divisors of $299$ are: $$1,\ 13,\ 23,\ 299$$ Thus, $$n-1\in\{1,13,23,299\}$$ which gives $$n\in\{2,14,24,300\}$$ Given $15\le n\le 50$, the only possible value is $$n=24$$ 3. **Find the common difference** Using $$299=(n-1)d$$ $$299=23d$$ $$d=13$$ 4. **Find $S_n$** Sum of first $n$ terms of an A.P. is $$S_n=\frac{n}{2}(a_1+a_n)$$ So, $$S_{24}=\frac{24}{2}(1+300)=12\cdot 301=3612$$ 5. **Interpret the required pair** The pair asked is $(S_{n-4}, a_{n-4})$. Since $n=24$, we need $$(S_{20}, a_{20})$$ 6. **Find $a_{20}$** $$a_{20}=a_1+19d=1+19\cdot 13=1+247=248$$ 7. **Find $S_{20}$** $$S_{20}=\frac{20}{2}(a_1+a_{20})=10(1+248)=10\cdot 249=2490$$ So, $$(S_{n-4}, a_{n-4})=(2490,248)$$ 8. **Match with the options** This is **Option C**.