Sequences and Series — JEE MCQs with answers

Answer: A — Option A is correct: NCERT prints exactly this: '2, 4, 8, 16, 32, ..., 1024. These numbers form what we call a sequence.' Option B is wrong: This stops at $2^9 = 512$, which is the ninth generation; the tenth is $2^{10} = 1024$, the number NCERT actually prints. Option C is wrong: This adds 2 ancestors per generation instead of doubling them; it would end at 20, whereas NCERT prints 1024 for the tenth generation. Option D is wrong: This starts the count at the person themselves ($1$), which halves every entry and ends at $2^9 = 512$; NCERT starts at the two parents, $a_1 = 2$, and ends at $a_{10} = 1024$.

Answer: B — Option B is correct: arithmetic mean, geometric mean and the relationship between A.M. and G.M. - three entries, printed one after another. Option A is wrong because this counts only “arithmetic mean” and stops at the first comma. Option C is wrong because means are named explicitly; the sentence lists them immediately after “more about A.P.;”. Option D is wrong because this counts the two means but overlooks the third entry, the relationship between A.M. and G.M., which is listed separately.

Answer: B — Option B is correct: Section 8.4.1 calls $a+ar+ar^2+\ldots+ar^{n-1}+\ldots$ an infinite geometric series, and that is as far as it goes: Section 8.4.2 sums only to $n$ terms, and no formula for an infinite sum appears anywhere in the Chapter. Option A (It names them and gives their sum as $\frac{a}{1-r}$.) is wrong because no such formula appears in this Chapter; Section 8.4.2 derives only the sum to $n$ terms. Option C (It never mentions infinite geometric series at all.) is wrong because the phrase is printed in Section 8.4.1, which names finite and infinite geometric series side by side. Option D (It names them and states that they may be summed when $|r|<1$.) is wrong because the condition $|r|<1$ is nowhere in this Chapter; the only condition attached to a sum formula here is $r\ne1$.

Answer: D — Option D is correct. Putting $\dfrac{a+b}{2}-\sqrt{ab}$ over the common denominator $2$ turns $\sqrt{ab}$ into $\dfrac{2\sqrt{ab}}{2}$. For $a=9,\ b=25$ it gives $\dfrac{34-30}{2}=2$, which is $A-G=17-15$. Option A is wrong. This subtracts the wrong way round and also forgets to double $\sqrt{ab}$. For $a=9,\ b=25$ it gives $\dfrac{15-34}{2}=-9.5$, which is negative, not $A-G=2$. Option B is wrong. The subtraction is taken the wrong way round. For $a=9,\ b=25$ it gives $\dfrac{30-34}{2}=-2$, the negative of $A-G$, so it could not be the non-negative quantity the section is heading towards. Option C is wrong. This forgets to double $\sqrt{ab}$ when putting it over the denominator $2$. For $a=9,\ b=25$ it gives $\dfrac{34-15}{2}=9.5$, not $2$.

Answer: C — Option C is correct. Putting $n = 1, 2, 3$: $a_1 = 2(1) + 5 = 7$, $a_2 = 2(2) + 5 = 9$, $a_3 = 2(3) + 5 = 11$, which is what the Solution prints. Option A is wrong. These are $a_2, a_3, a_4$ -- the list has been started one step too late. Option B is wrong. These are the values at $n = 0, 1, 2$; the Solution substitutes $n = 1, 2, 3$, because the first term of a sequence is $a_1$. Option D is wrong. Dropping the constant leaves $2n$, whose first three values are $2, 4, 6$; the given rule adds $5$ to each.

Answer: D — Option D is correct: $\frac{x^{5}}{x^{3}}=x^{2}$ and $\frac{x^{7}}{x^{5}}=x^{2}$. The value $x$ would give a second term of $x^{4}$, not $x^{5}$. The value $x^{3}$ is the first term, and using it as the ratio would give a second term of $x^{6}$. The value $2$ comes from reading the difference of the exponents as the ratio itself, which would give a second term of $2x^{3}$.

Answer: B — Option B is correct. $a_1 = \dfrac{1-3}{4} = -\dfrac{1}{2}$, $a_2 = \dfrac{2-3}{4} = -\dfrac{1}{4}$, $a_3 = \dfrac{3-3}{4} = 0$ -- the three values the Solution prints. Option A is wrong. These come from computing $\dfrac{3-n}{4}$, that is, from subtracting in the wrong order; the rule has $n$ first. Option C is wrong. These are the values at $n = 0, 1, 2$; the first term of a sequence is $a_1$, not $a_0$. Option D is wrong. These are the numerators $n - 3$ with the division by $4$ forgotten.

Answer: C — Option C is correct. Let the G.P. have first term $\alpha$ and ratio $\rho$, so $a=\alpha\rho^{p-1}$, $b=\alpha\rho^{q-1}$, $c=\alpha\rho^{r-1}$. The exponents of $\alpha$ add to $(q-r)+(r-p)+(p-q)=0$, and those of $\rho$ add to $(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)=0$ as well, so the product is $1$. This is the statement printed on page 146. Option A is wrong. The exponents are right but the value is not: they cancel completely, so the product is $1$, not $abc$. For the G.P. with $\alpha=3,\ \rho=2$ and $p,q,r=2,3,5$ we get $a,b,c=6,12,48$, the product is $1$ and $abc=3456$. Option B is wrong. The exponents have been attached to the wrong bases and the value changed to $abc$. For $a,b,c=6,12,48$ (positions $2,3,5$) the left side is $6^{-1}\cdot12^{-2}\cdot48^{3}=128$, not $abc=3456$. Option D is wrong. The exponents have been attached to the wrong bases. The exponent on $a$ must be the difference of the other two positions, $q-r$, for the powers of the ratio to cancel; for $a,b,c=6,12,48$ (positions $2,3,5$) this left side is $128$, not $1$.

Answer: B — Option B is correct: NCERT writes 'The nth term is the number at the nth position of the sequence and is denoted by $a_n$.' Option A is wrong: That holds only for the sequence 1, 2, 3, ...; in the ancestor sequence $a_3 = 8$, not 3. Option C is wrong: The number of terms is what decides whether a sequence is finite; $a_n$ is one particular term. Option D is wrong: Section 8.2 defines no sums at all; $a_n$ is a single term.

Answer: C — Option C is correct: the chapter carries a numbered sub-section 8.4.3 headed “Geometric Mean (G.M.)”, which defines it for two positive numbers. Option A is wrong because no numbered section of the chapter develops A.P.; it is referred to only as prior knowledge. Option B is wrong because no such result appears anywhere in the chapter. Option D is wrong because no such result appears anywhere in the chapter.