Let a1, a2, ......., a10 be an AP with common difference - 3 and b1, b2, ........., b10 be a GP with common ratio 2. Let ck = ak + bk, k = 1, 2, ......, 10. If c2 = 12 and c3 = 13, then \sum\limits_k = 1^10 c_k is equal to \underline\hspace2cm .

Sequences and Series PYQ — JEE Main 2021 · 26 Aug · Shift 2 · Q42 with Solution

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Correct answer: 2021

For the AP with common difference $-3$ : $a_k=a_1+(k-1)(-3)=a_1-3(k-1).$

For the GP with common ratio $2$ : $b_k=b_1\cdot 2^{k-1}.$

Given $c_k=a_k+b_k.$

So, $c_k=a_1-3(k-1)+b_1 2^{k-1}.$

For $k=2$ :

\qquad b_2=2b_1.$$ Hence, $$c_2=(a_1-3)+2b_1=12$$ $$a_1+2b_1=15 \quad ...(1)$$ For $k=3$: $$a_3=a_1-6, \qquad b_3=4b_1.$$ Hence, $$c_3=(a_1-6)+4b_1=13$$ $$a_1+4b_1=19 \quad ...(2)$$ Subtract (1) from (2): $$2b_1=4 \Rightarrow b_1=2.$$ Substitute into (1): $$a_1+2(2)=15$$ $$a_1=11.$$ --- 3. **Find the sum of the AP and GP separately** ### Sum of AP Here $a_1=11$, $d=-3$, $n=10$. $$S_A=\frac{10}{2}\left[2(11)+(10-1)(-3)\right]$$ $$S_A=5(22-27)=5(-5)=-25.$$ ### Sum of GP Here $b_1=2$, $r=2$, $n=10$. $$S_B=b_1\frac{r^{10}-1}{r-1}$$ $$S_B=2\cdot \frac{2^{10}-1}{2-1}=2(1024-1)=2\cdot 1023=2046.$$ --- 4. **Find the required sum** Since $c_k=a_k+b_k$, $$\sum_{k=1}^{10} c_k=\sum_{k=1}^{10} a_k+\sum_{k=1}^{10} b_k.$$ Therefore, $$\sum_{k=1}^{10} c_k=-25+2046=2021.$$ --- 5. **Comparison with stored answer** Derived answer = $2021$. Stored correct answer = $2021$. So, the answer agrees with the stored correct answer.

View written solutionFree

Correct answer: 2021

For the AP with common difference $-3$ : $a_k=a_1+(k-1)(-3)=a_1-3(k-1).$

For the GP with common ratio $2$ : $b_k=b_1\cdot 2^{k-1}.$

Given $c_k=a_k+b_k.$

So, $c_k=a_1-3(k-1)+b_1 2^{k-1}.$

For $k=2$ :

\qquad b_2=2b_1.$$ Hence, $$c_2=(a_1-3)+2b_1=12$$ $$a_1+2b_1=15 \quad ...(1)$$ For $k=3$: $$a_3=a_1-6, \qquad b_3=4b_1.$$ Hence, $$c_3=(a_1-6)+4b_1=13$$ $$a_1+4b_1=19 \quad ...(2)$$ Subtract (1) from (2): $$2b_1=4 \Rightarrow b_1=2.$$ Substitute into (1): $$a_1+2(2)=15$$ $$a_1=11.$$ --- 3. **Find the sum of the AP and GP separately** ### Sum of AP Here $a_1=11$, $d=-3$, $n=10$. $$S_A=\frac{10}{2}\left[2(11)+(10-1)(-3)\right]$$ $$S_A=5(22-27)=5(-5)=-25.$$ ### Sum of GP Here $b_1=2$, $r=2$, $n=10$. $$S_B=b_1\frac{r^{10}-1}{r-1}$$ $$S_B=2\cdot \frac{2^{10}-1}{2-1}=2(1024-1)=2\cdot 1023=2046.$$ --- 4. **Find the required sum** Since $c_k=a_k+b_k$, $$\sum_{k=1}^{10} c_k=\sum_{k=1}^{10} a_k+\sum_{k=1}^{10} b_k.$$ Therefore, $$\sum_{k=1}^{10} c_k=-25+2046=2021.$$ --- 5. **Comparison with stored answer** Derived answer = $2021$. Stored correct answer = $2021$. So, the answer agrees with the stored correct answer.