Let a_1, a_2, \ldots, a_n be in A.P. If a_5=2 a_7 and a_11=18 , then 12(\frac1\sqrta_10+\sqrta_11+\frac1\sqrta_11+\sqrta_12+\ldots+\frac1\sqrta_17+\sqrta_18) is equal to \underline\hspace2cm .

Sequences and Series PYQ — JEE Main 2023 · 31 Jan · Shift 1 · Q44 with Solution

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Correct answer: 8

Write the A.P. terms in standard form

Let the arithmetic progression be $a_n=a+(n-1)d.$

Then:

\qquad a_7=a+6d.$$ Given: $$a_5=2a_7$$ so, $$a+4d=2(a+6d).$$ Simplifying, $$a+4d=2a+12d$$ $$0=a+8d$$ $$a=-8d.$$ 2. **Use the condition $a_{11}=18$** Now, $$a_{11}=a+10d=18.$$ Substitute $a=-8d$: $$-8d+10d=18$$ $$2d=18$$ $$d=9.$$ Hence, $$a=-72.$$ Therefore, $$a_n=-72+9(n-1)=9n-81=9(n-9).$$ 3. **Find the required terms** So, $$a_{10}=9, \quad a_{11}=18, \quad a_{12}=27, \quad ext{and in general } a_n=9(n-9).$$ Thus, $$\sqrt{a_n}=\sqrt{9(n-9)}=3\sqrt{n-9}.$$ 4. **Evaluate the sum** We need $$12\sum_{k=10}^{17} \frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}.$$ Using $a_k=9(k-9)$, $$\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} =\frac{1}{3\sqrt{k-9}+3\sqrt{k-8}} =\frac{1}{3(\sqrt{k-9}+\sqrt{k-8})}.$$ So the whole expression becomes $$12\sum_{k=10}^{17}\frac{1}{3(\sqrt{k-9}+\sqrt{k-8})} =4\sum_{k=10}^{17}\frac{1}{\sqrt{k-9}+\sqrt{k-8}}.$$ Let $m=k-9$. Then as $k$ goes from $10$ to $17$, $m$ goes from $1$ to $8$: $$4\sum_{m=1}^{8}\frac{1}{\sqrt{m}+\sqrt{m+1}}.$$ Now rationalize: $$\frac{1}{\sqrt{m}+\sqrt{m+1}} =\frac{\sqrt{m+1}-\sqrt{m}}{(m+1)-m} =\sqrt{m+1}-\sqrt{m}.$$ Hence the sum telescopes: $$4\sum_{m=1}^{8}(\sqrt{m+1}-\sqrt{m}) =4(\sqrt{9}-\sqrt{1}) =4(3-1)=8.$$ 5. **Final answer** $$\boxed{8}$$ 6. **Comparison with stored answer** Stored correct answer = $8$. My derived answer matches it.

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View written solutionFree

Correct answer: 8

Write the A.P. terms in standard form

Let the arithmetic progression be $a_n=a+(n-1)d.$

Then:

\qquad a_7=a+6d.$$ Given: $$a_5=2a_7$$ so, $$a+4d=2(a+6d).$$ Simplifying, $$a+4d=2a+12d$$ $$0=a+8d$$ $$a=-8d.$$ 2. **Use the condition $a_{11}=18$** Now, $$a_{11}=a+10d=18.$$ Substitute $a=-8d$: $$-8d+10d=18$$ $$2d=18$$ $$d=9.$$ Hence, $$a=-72.$$ Therefore, $$a_n=-72+9(n-1)=9n-81=9(n-9).$$ 3. **Find the required terms** So, $$a_{10}=9, \quad a_{11}=18, \quad a_{12}=27, \quad ext{and in general } a_n=9(n-9).$$ Thus, $$\sqrt{a_n}=\sqrt{9(n-9)}=3\sqrt{n-9}.$$ 4. **Evaluate the sum** We need $$12\sum_{k=10}^{17} \frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}.$$ Using $a_k=9(k-9)$, $$\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} =\frac{1}{3\sqrt{k-9}+3\sqrt{k-8}} =\frac{1}{3(\sqrt{k-9}+\sqrt{k-8})}.$$ So the whole expression becomes $$12\sum_{k=10}^{17}\frac{1}{3(\sqrt{k-9}+\sqrt{k-8})} =4\sum_{k=10}^{17}\frac{1}{\sqrt{k-9}+\sqrt{k-8}}.$$ Let $m=k-9$. Then as $k$ goes from $10$ to $17$, $m$ goes from $1$ to $8$: $$4\sum_{m=1}^{8}\frac{1}{\sqrt{m}+\sqrt{m+1}}.$$ Now rationalize: $$\frac{1}{\sqrt{m}+\sqrt{m+1}} =\frac{\sqrt{m+1}-\sqrt{m}}{(m+1)-m} =\sqrt{m+1}-\sqrt{m}.$$ Hence the sum telescopes: $$4\sum_{m=1}^{8}(\sqrt{m+1}-\sqrt{m}) =4(\sqrt{9}-\sqrt{1}) =4(3-1)=8.$$ 5. **Final answer** $$\boxed{8}$$ 6. **Comparison with stored answer** Stored correct answer = $8$. My derived answer matches it.

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