Arithmetic Progression — JEE Main Maths PYQ MCQs with Solutions

Solution: Given $a_n = 3n + 2$. First term $a_1 = 3(1) + 2 = 5$. Second term $a_2 = 3(2) + 2 = 8$. Common difference $d = 8 - 5 = 3$. Sum formula: $S_n = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(2(5) + (n-1)3) = \frac{n}{2}(10 + 3n - 3) = \frac{n(3n+7)}{2}$.

Solution: Sum formula: $S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}(2 + 14) = 8n$. Given $S_n = 128$, so $8n = 128 \Rightarrow n = 16$.

Solution: Natural numbers between 100 and 200 divisible by 3: 102, 105, 108, ..., 198. This is an AP with $a = 102$, $d = 3$, $l = 198$. Number of terms: $n = \frac{198 - 102}{3} + 1 = 32 + 1 = 33$. Sum: $S = \frac{33}{2}(102 + 198) = \frac{33 \times 300}{2} = 4950$.

Solution: Let $S_m = n$ and $S_n = m$. For an AP, $S_k = \frac{k}{2}[2a + (k-1)d]$. From the given conditions and algebraic manipulation, $S_{m+n} = -(m+n)$.

Solution: The arithmetic mean (AM) between two numbers $a$ and $b$ is the middle term of the AP formed by $a$, AM, $b$. If this middle term is $x$, then $x - a = b - x$ (equal common differences), so $2x = a + b$, hence $x = \frac{a+b}{2}$.

Solution: If $a$, $b$, $c$ are in AP, then $b - a = d$ (common difference) and $c - b = d$. Therefore $b - a = c - b$.

Solution: Given $a = 1$, $d = 4$. Sum formula: $S_n = \frac{n}{2}[2(1) + (n-1)4] = \frac{n}{2}(4n - 2) = n(2n - 1)$. We need $n(2n-1) > 100$. For $n = 8$: $8(16-1) = 8 \times 15 = 120 > 100$. So $n = 8$.

Solution: Let the AP have first term $A$ and common difference $D$. Then $a = A + (p-1)D$, $b = A + (q-1)D$, $c = A + (r-1)D$. The expression $(q-r)a + (r-p)b + (p-q)c$ expands and simplifies to 0.

Solution: The first $n$ odd numbers are $1, 3, 5, \ldots, (2n-1)$. This is an AP with $a = 1$, $d = 2$, and $l = 2n-1$. Sum = $\frac{n}{2}(1 + 2n - 1) = \frac{n}{2}(2n) = n^2$.

Solution: Given $S_n = 2n^2 + 3n$. The $n$-th term $a_n = S_n - S_{n-1}$ for $n \geq 2$. This gives $a_n = 4n + 1$. Checking for $n = 1$: $a_1 = S_1 = 5 = 4(1) + 1$. So $a_n = 4n + 1$.

Solution: Let the three numbers in AP be $a-d$, $a$, $a+d$. Sum: $3a = 24 \Rightarrow a = 8$. Product: $a(a^2 - d^2) = 8(64 - d^2) = 440 \Rightarrow d^2 = 9 \Rightarrow d = 3$. Numbers are $5, 8, 11$.

Solution: From $S_n = 3n^2 - 2n$, we get $a_1 = S_1 = 1$ and $a_2 = S_2 - S_1 = 8 - 1 = 7$. Common difference $d = a_2 - a_1 = 6$.

Solution: Given $a_n = 5n - 3$. First term $a_1 = 2$. 20th term $a_{20} = 97$. Sum of first 20 terms: $S_{20} = \frac{20}{2}(2 + 97) = 10 \times 99 = 990$.

Solution: For two APs with $\frac{S_n}{S_n'} = \frac{7n+1}{4n+27}$, the ratio of $n$-th terms is $\frac{a_n}{b_n} = \frac{S_{2n-1}}{S_{2n-1}'} = \frac{7(2n-1)+1}{4(2n-1)+27} = \frac{14n - 6}{8n + 23}$.

Solution: $(a_1 - a_2) + (a_3 - a_4) + \ldots + (a_{2n-1} - a_{2n}) = (-d) + (-d) + \ldots + (-d) = -nd$ (there are $n$ pairs).

Solution: First term $a = 3$, common difference $d = 4$, last term $l = 399$. Using $l = a + (n-1)d$: $399 = 3 + 4(n-1) \Rightarrow 396 = 4(n-1) \Rightarrow n = 100$.

Solution: Let common difference be $D$. Then $b = a + D$, $c = a + 2D$, $d = a + 3D$. Now $a - 3b + 3c - d = a - 3(a+D) + 3(a+2D) - (a+3D) = 0$.

Solution: The series $1, 2, 3, \ldots, n$ is an AP with $a = 1$, $d = 1$, $l = n$. Sum = $\frac{n}{2}(1 + n) = \frac{n(n+1)}{2}$.

Solution: Using $a_n = a + (n-1)d$: $95 = 10 + 5(n-1) \Rightarrow 85 = 5(n-1) \Rightarrow n = 18$.

Solution: $T_n = 2n + 1$. So $T_{10} = 2(10) + 1 = 21$.

Solution: Using the same approach as for similar problems, we derive that $S_{p+q} = -(p+q)$.

Solution: Using the formula $S = \frac{20(21)(41)}{6} = \frac{17220}{6} = 2870$.

Solution: Let the AP have first term $a$ and common difference $d$. From $a_m = \frac{1}{n}$ and $a_n = \frac{1}{m}$, we get $(m-n)d = \frac{m-n}{mn}$, so $d = \frac{1}{mn}$. Then $a = \frac{1}{mn}$. Thus $a_{mn} = \frac{1}{mn} + (mn-1) \cdot \frac{1}{mn} = 1$.

Solution: If $S_n = 0$, then $\frac{n}{2}[2a + (n-1)d] = 0 \Rightarrow 2a + (n-1)d = 0$. This shows $a$ and $d$ must have opposite signs.

Solution: Given $S_{2n} = 3S_n$. We have $S_n = \frac{n}{2}[2a + (n-1)d]$ and $S_{2n} = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d]$. From $S_{2n} = 3S_n$: $n[2a + (2n-1)d] = 3 \cdot \frac{n}{2}[2a + (n-1)d] \Rightarrow 2[2a + (2n-1)d] = 3[2a + (n-1)d] \Rightarrow 4a + (4n-2)d = 6a + (3n-3)d \Rightarrow (4n-2-3n+3)d = 2a \Rightarrow (n+1)d = 2a$. So $a = \frac{(n+1)d}{2}$. Then $a_n = a + (n-1)d = \frac{(n+1)d}{2} + (n-1)d = \frac{(n+1)d + 2(n-1)d}{2} = \frac{(3n-1)d}{2}$. And $S_{2n} = 3S_n = 3 \cdot \frac{n}{2}[2a + (n-1)d] = \frac{3n}{2}[(n+1)d + (n-1)d] = \frac{3n}{2} \cdot 2nd = 3n^2d$. Thus $\frac{a_n}{S_{2n}} = \frac{(3n-1)d/2}{3n^2d} = \frac{3n-1}{6n^2}$. For large $n$ this approaches $\frac{1}{2n}$. Actually let me recalculate more carefully for the exact ratio.