Suppose four distinct positive numbers a_1,\,a_2\,,\,a_3,\,a_4\, are in G.P. Let b_1 = a_1,b_2 = b_1 + a_2,\,b_3 = b_2 + a_3\,\,\,\,\,and\,\,\,b_4 = b_3 + a_4 . STATEMENT-1: The numbers b_1,\,b_2\,,\,b_3,\,b_4\, are neither in A.P. nor in G.P. and STATEMENT-2 The numbers b_1,\,b_2\,,\,b_3,\,b_4\, are in H.P.

STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

STATEMENT-1 is False, STATEMENT-2 is True

Sequences and Series PYQ — JEE Advanced 2008 · Shift 2 · Q38 with Solution

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Correct answer: C

Write the G.P. terms in standard form

Let the four distinct positive numbers in G.P. be

\quad a_2=ar, \quad a_3=ar^2, \quad a_4=ar^3$$ with $a>0$, $r>0$, and since the terms are distinct, $r\neq 1$. 2. **Form the partial sums** Given $$b_1=a_1, \quad b_2=a_1+a_2, \quad b_3=a_1+a_2+a_3, \quad b_4=a_1+a_2+a_3+a_4.$$ So, $$b_1=a,$$ $$b_2=a(1+r),$$ $$b_3=a(1+r+r^2),$$ $$b_4=a(1+r+r^2+r^3).$$ --- 3. **Check whether $b_1,b_2,b_3,b_4$ are in A.P.** For A.P., consecutive differences must be equal. Compute: $$b_2-b_1=ar,$$ $$b_3-b_2=ar^2,$$ $$b_4-b_3=ar^3.$$ These are equal only if $$ar=ar^2=ar^3.$$ Since $a>0$, this gives $$r=r^2=r^3 \implies r=1,$$ which is not allowed because the terms are distinct. Hence, $b_1,b_2,b_3,b_4$ are **not in A.P.** --- 4. **Check whether $b_1,b_2,b_3,b_4$ are in G.P.** For G.P., ratios of consecutive terms must be equal. First, $$\frac{b_2}{b_1}=\frac{a(1+r)}{a}=1+r.$$ Next, $$\frac{b_3}{b_2}=\frac{1+r+r^2}{1+r}.$$ For G.P., we need $$1+r=\frac{1+r+r^2}{1+r}.$$ Multiplying, $$(1+r)^2=1+r+r^2.$$ Expanding, $$1+2r+r^2=1+r+r^2 \implies r=0,$$ which is impossible since all terms are positive. Therefore, the sequence is **not in G.P.** So **STATEMENT-1 is true**. --- 5. **Check whether $b_1,b_2,b_3,b_4$ are in H.P.** A sequence is in H.P. if its reciprocals are in A.P. Now, $$b_1=a, \quad b_2=a(1+r), \quad b_3=a(1+r+r^2), \quad b_4=a(1+r+r^2+r^3).$$ Their reciprocals are $$\frac1{b_1}=\frac1a, \quad \frac1{b_2}=\frac1{a(1+r)}, \quad \frac1{b_3}=\frac1{a(1+r+r^2)}, \quad \frac1{b_4}=\frac1{a(1+r+r^2+r^3)}.$$ To test A.P. of reciprocals, check whether $$\frac{2}{b_2}=\frac1{b_1}+\frac1{b_3}.$$ That is, $$\frac{2}{a(1+r)}=\frac1a+\frac1{a(1+r+r^2)}.$$ Multiply by $a$: $$\frac{2}{1+r}=1+\frac1{1+r+r^2}.$$ Right side becomes $$1+\frac1{1+r+r^2}=\frac{2+r+r^2}{1+r+r^2}.$$ So we need $$\frac{2}{1+r}=\frac{2+r+r^2}{1+r+r^2}.$$ Cross-multiplying, $$2(1+r+r^2)=(1+r)(2+r+r^2).$$ LHS: $$2+2r+2r^2$$ RHS: $$2+3r+2r^2+r^3$$ Equating: $$2+2r+2r^2=2+3r+2r^2+r^3$$ $$0=r+r^3=r(1+r^2).$$ Since $r>0$, this is impossible. Hence reciprocals are not in A.P., so $b_1,b_2,b_3,b_4$ are **not in H.P.** Thus **STATEMENT-2 is false**. --- 6. **Conclusion** - STATEMENT-1: True - STATEMENT-2: False Therefore, the correct option is: $$\boxed{\text{C}}$$

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View written solutionFree

Correct answer: C

Write the G.P. terms in standard form

Let the four distinct positive numbers in G.P. be

\quad a_2=ar, \quad a_3=ar^2, \quad a_4=ar^3$$ with $a>0$, $r>0$, and since the terms are distinct, $r\neq 1$. 2. **Form the partial sums** Given $$b_1=a_1, \quad b_2=a_1+a_2, \quad b_3=a_1+a_2+a_3, \quad b_4=a_1+a_2+a_3+a_4.$$ So, $$b_1=a,$$ $$b_2=a(1+r),$$ $$b_3=a(1+r+r^2),$$ $$b_4=a(1+r+r^2+r^3).$$ --- 3. **Check whether $b_1,b_2,b_3,b_4$ are in A.P.** For A.P., consecutive differences must be equal. Compute: $$b_2-b_1=ar,$$ $$b_3-b_2=ar^2,$$ $$b_4-b_3=ar^3.$$ These are equal only if $$ar=ar^2=ar^3.$$ Since $a>0$, this gives $$r=r^2=r^3 \implies r=1,$$ which is not allowed because the terms are distinct. Hence, $b_1,b_2,b_3,b_4$ are **not in A.P.** --- 4. **Check whether $b_1,b_2,b_3,b_4$ are in G.P.** For G.P., ratios of consecutive terms must be equal. First, $$\frac{b_2}{b_1}=\frac{a(1+r)}{a}=1+r.$$ Next, $$\frac{b_3}{b_2}=\frac{1+r+r^2}{1+r}.$$ For G.P., we need $$1+r=\frac{1+r+r^2}{1+r}.$$ Multiplying, $$(1+r)^2=1+r+r^2.$$ Expanding, $$1+2r+r^2=1+r+r^2 \implies r=0,$$ which is impossible since all terms are positive. Therefore, the sequence is **not in G.P.** So **STATEMENT-1 is true**. --- 5. **Check whether $b_1,b_2,b_3,b_4$ are in H.P.** A sequence is in H.P. if its reciprocals are in A.P. Now, $$b_1=a, \quad b_2=a(1+r), \quad b_3=a(1+r+r^2), \quad b_4=a(1+r+r^2+r^3).$$ Their reciprocals are $$\frac1{b_1}=\frac1a, \quad \frac1{b_2}=\frac1{a(1+r)}, \quad \frac1{b_3}=\frac1{a(1+r+r^2)}, \quad \frac1{b_4}=\frac1{a(1+r+r^2+r^3)}.$$ To test A.P. of reciprocals, check whether $$\frac{2}{b_2}=\frac1{b_1}+\frac1{b_3}.$$ That is, $$\frac{2}{a(1+r)}=\frac1a+\frac1{a(1+r+r^2)}.$$ Multiply by $a$: $$\frac{2}{1+r}=1+\frac1{1+r+r^2}.$$ Right side becomes $$1+\frac1{1+r+r^2}=\frac{2+r+r^2}{1+r+r^2}.$$ So we need $$\frac{2}{1+r}=\frac{2+r+r^2}{1+r+r^2}.$$ Cross-multiplying, $$2(1+r+r^2)=(1+r)(2+r+r^2).$$ LHS: $$2+2r+2r^2$$ RHS: $$2+3r+2r^2+r^3$$ Equating: $$2+2r+2r^2=2+3r+2r^2+r^3$$ $$0=r+r^3=r(1+r^2).$$ Since $r>0$, this is impossible. Hence reciprocals are not in A.P., so $b_1,b_2,b_3,b_4$ are **not in H.P.** Thus **STATEMENT-2 is false**. --- 6. **Conclusion** - STATEMENT-1: True - STATEMENT-2: False Therefore, the correct option is: $$\boxed{\text{C}}$$

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