08. Arithmetic Progression (AP) Previous Year Questions

a = 5, l = 50, Sn = 440.
(a) Sn = n/2 (a + l) ⇒ 440 = n/2 (5 + 50) = 55n/2 ⇒ n = 16 terms. [1.5 Marks]
(b) l = a + (n − 1)d ⇒ 50 = 5 + 15d ⇒ 15d = 45 ⇒ d = 3. [1.5 Marks]

Correct Option: (a) (¹/_a + 6)
First term a₁ = ¹/_a, d = 1.
T₇ = a₁ + 6d = ¹/_a + 6. [1 Mark]

T_n = 42 − 6n.
(a) First term a = T₁ = 42 − 6 = 36. T₂ = 30 ⇒ common difference d = 30 − 36 = −6. [1.5 Marks]
(b) S₂₅ = (25/2)[2(36) + 24(−6)] = (25/2)[72 − 144] = 25 × (−36) = −900. [1.5 Marks]

Given $T_n = 2n + 5$. For $n = 10$:
$T_{10} = 2(10) + 5 = 20 + 5 = 25$.
Correct Option: (c) [1 Mark]

Let the first term be $a$ and common difference be $d$.
$T_5 = a + 4d = 4$ — (Eq. 1)
$T_9 = a + 8d = -12$ — (Eq. 2)

Subtract Eq. 1 from Eq. 2
$(a + 8d) - (a + 4d) = -12 - 4 \implies 4d = -16 \implies \mathbf{d = -4}$. (Common Difference)1 Mark

Find First Term
Substitute $d = -4$ into Eq. 1:
$a + 4(-4) = 4 \implies a - 16 = 4 \implies \mathbf{a = 20}$.1 Mark

Sum of 16 Terms
$S_n = \frac{n}{2}[2a + (n - 1)d]$
$S_{16} = \frac{16}{2}[2(20) + (16 - 1)(-4)] = 8[40 + 15(-4)] = 8[40 - 60] = 8(-20) = \mathbf{-160}$.1 Mark

First term $a = 15$. Common difference $d = 30 - 15 = 15$.

Part 1 $n^\text{th}$ term formula: $T_n = a + (n - 1)d = 300$
$15 + (n - 1)15 = 300 \implies 15n = 300 \implies \mathbf{n = 20}$.
Therefore, the $20^\text{th}$ term is 300.1.5 Marks

Part 2 Sum of $n$ terms $S_n = \frac{n}{2}(a + l)$ where last term $l = 300$:
$S_{20} = \frac{20}{2}(15 + 300) = 10 \times 315 = \mathbf{3,150}$.1.5 Marks

Step 1 Let the first term be $a$ and common difference be $d$.
$T_6 = a + 5d$. Given $T_6 = 4a$:
$$a + 5d = 4a \implies 5d = 3a \implies a = \frac{5}{3}d$$ [1 Mark]
Step 2 Given $S_6 = 75$:
$$S_6 = \frac{6}{2}[2a + (6 - 1)d] = 3(2a + 5d) = 75 \implies 2a + 5d = 25$$ [1 Mark]
Step 3 Substitute $5d = 3a$:
$$2a + 3a = 25 \implies 5a = 25 \implies \mathbf{a = 5}$$
Then $5d = 3(5) = 15 \implies \mathbf{d = 3}$.
Answer: First term $a = 5$, common difference $d = 3$. [1 Mark]

Step 1 Set up linear equations in $a$ and $d$:
$T_4 = a + 3d = 8$   ...(i)
$T_6 = a + 5d = 14$   ...(ii) [1 Mark]
Step 2 Subtract (i) from (ii):
$(a + 5d) - (a + 3d) = 14 - 8 \implies 2d = 6 \implies \mathbf{d = 3}$.
$\therefore \text{Common difference } d = \mathbf{3}$. [1 Mark]
Step 3 Substitute $d = 3$ into (i):
$a + 3(3) = 8 \implies a + 9 = 8 \implies \mathbf{a = -1}$.
$\therefore \text{First term } a = \mathbf{-1}$. [1 Mark]
Step 4 Find sum of the first 20 terms ($S_{20}$):
$S_n = \frac{n}{2}[2a + (n - 1)d]$
$S_{20} = \frac{20}{2}[2(-1) + (20 - 1)(3)] = 10[-2 + 19(3)] = 10[-2 + 57] = 10(55) = \mathbf{550}$. [1 Mark]

Step 1 Let the three consecutive terms of the A.P. be $(a - d)$, $a$, and $(a + d)$.
Sum of the three terms $= (a - d) + a + (a + d) = 3a = 42 \implies \mathbf{a = 14}$. [1 Mark]
Step 2 Product of the first and third terms $= 52$:
$(a - d)(a + d) = a^2 - d^2 = 52$.
Substitute $a = 14$:
$14^2 - d^2 = 52 \implies 196 - d^2 = 52 \implies d^2 = 196 - 52 = 144 \implies \mathbf{d = \pm 12}$. [1 Mark]
Step 3 Determine the terms and solutions:
Case 1 ($d = 12$):
First term $T_1 = a - d = 14 - 12 = \mathbf{2}$.
Common difference $d = \mathbf{12}$. (Terms: $2, 14, 26$)
Case 2 ($d = -12$):
First term $T_1 = a - d = 14 - (-12) = \mathbf{26}$.
Common difference $d = \mathbf{-12}$. (Terms: $26, 14, 2$)
Both solutions are completely valid. [1 Mark]

Step 1 (a) Simplify surd terms:
$T_1 = \sqrt{2}$, $T_2 = \sqrt{8} = 2\sqrt{2}$, $T_3 = \sqrt{18} = 3\sqrt{2}$.
$T_2 - T_1 = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$, and $T_3 - T_2 = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$.
Since the difference between consecutive terms is constant, it is an **Arithmetic Progression (A.P.)** with $a = \sqrt{2}$ and $d = \sqrt{2}$. [2 Marks]
Step 2 (b) Calculate sum of first 10 terms ($S_{10}$):
$S_{10} = \frac{10}{2}[2a + (10 - 1)d] = 5[2\sqrt{2} + 9\sqrt{2}] = 5(11\sqrt{2}) = 55\sqrt{2}$.
Given $S_{10} = p\sqrt{2} \implies \mathbf{p = 55}$. [2 Marks]

Step 1 Condition for consecutive terms $a, b, c$ in A.P.:
$$b - a = c - b \iff 2b = a + c$$ [1 Mark]
Step 2 Substitute the given expressions:
$$2(2k + 1) = (k - 3) + (4k + 3)$$ $$4k + 2 = 5k$$ [1 Mark]
Step 3 Solve for $k$:
$$5k - 4k = 2 \implies \mathbf{k = 2}$$ [1 Mark]
Check: Terms are $2 - 3 = -1$, $2(2) + 1 = 5$, and $4(2) + 3 = 11$. Common difference $d = 6$.
Answer: $k = 2$.

Step 1 Let first term be $a$ and common difference be $d$.
$$T_4 = a + 3d = 22 \quad \text{--- (1)}$$ [1 Mark]
$$T_{15} = a + 14d = 66 \quad \text{--- (2)}$$ [1 Mark]
Step 2 Subtract (1) from (2):
$$(a + 14d) - (a + 3d) = 66 - 22 \implies 11d = 44 \implies \mathbf{d = 4}$$ [0.5 Mark]
Substitute $d = 4$ into (1):
$$a + 3(4) = 22 \implies a + 12 = 22 \implies \mathbf{a = 10}$$ [0.5 Mark]
Step 3 Sum of first 8 terms ($S_8$):
$$S_n = \frac{n}{2}[2a + (n - 1)d]$$ $$S_8 = \frac{8}{2}[2(10) + (8 - 1)4] = 4[20 + 28] = 4 \times 48 = \mathbf{192}$$ [1 Mark]
Answers: First term $a = 10$, Common difference $d = 4$, Sum of 8 terms $S_8 = 192$.