Progressions - Aptitude MCQ Questions and Solutions with Explanations

First term of AP = a = -19 and last term = l = 36
Number of terms = n = 12
$$\eqalign{ & {\text{Sum of AP}} = \frac{n}{2}\left( {a + l} \right) \cr & = \frac{{12}}{2}\left( { - 19 + 36} \right) \cr & = 17 \times 6 \cr & = 102 \cr} $$

$$\eqalign{ & {T_3} = a + 2d = - 9\,.....\,\left( 1 \right) \cr & {T_7} = a + 6d = 11\,.....\,\left( 2 \right) \cr & {\text{on solving}}\left( 1 \right)\,{\text{and}}\,\left( 2 \right) \cr & d = 5\,\& \,a = - 19 \cr & {T_{15}} = a + 14d \cr & \,\,\,\,\,\,\,\,\,\, = - 19 + 14\left( 5 \right) \cr & \,\,\,\,\,\,\,\,\,\, = 51 \cr} $$

$$\eqalign{ & {T_3} = a + 2d = - 13\,.....\,\left( 1 \right) \cr & {T_6} = a + 5d = - 4\,.....\,\left( 2 \right) \cr & {\text{on solving}}\left( 1 \right)\,{\text{and}}\,\left( 2 \right) \cr & d = 3\& a = - 19 \cr & {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr} $$
$${S_{12}} = \frac{{12}}{2}$$ $$\left[ {2\left( { - 19} \right) + \left( {12 - 1} \right)\left( 3 \right)} \right]$$
$$\eqalign{ & {S_{12}} = \left( 6 \right)\left[ { - 38 + 33} \right] \cr & {S_{12}} = - 30 \cr} $$

$$\eqalign{ & {T_3} = a + 2d = 13\,.....\,\left( 1 \right) \cr & {T_5} = a + 4d = 21\,.....\,\left( 2 \right) \cr & {\text{on solving}}\left( 1 \right)\,{\text{and}}\,\left( 2 \right) \cr & d = 4\& a = 5 \cr & {T_{13}} = a + 12d \cr & \,\,\,\,\,\,\,\,\, = 5 + 12\left( 4 \right) \cr & \,\,\,\,\,\,\,\,\, = 5 + 48 \cr & \,\,\,\,\,\,\,\,\, = 53 \cr} $$

T₃ = a + 2d = 13 ...... (1)
T₆ = a + 5d = -5 ...... (2)
on solving (1) and (2) d = -6 & a = 25
T₁₁ = a + 10d
      = 25 + 10(-6)
      = -35

Let the first term of an AP = a and the common difference = d
3^th term of AP = A₃ = a + 2d = -8 ......(1)
9^th term = A₉ = a + 8d = 10 ...... (2)
Subtracting equation (1) from (2), we get :
⇒ 8d - 2d = 10 - (-8)
⇒ 6d =18
⇒ d = $$\frac{{18}}{6}$$ = 3
Substituting it in equation (2),
⇒ a = 10 - 8(3)
      = 10 - 24
      = -14
∴ 16^th term = A₁₆ = a + 15d
= -14 + 15(3)
= -14 + 45
= 31

Arithmetic mean of 7, 5, 13, x and 9 = 10
$$\eqalign{ & \Rightarrow \frac{{7 + 5 + 13 + x + 9}}{5} = 10 \cr & \Rightarrow 34 + x = 10 \times 5 \cr & \Rightarrow 34 + x = 50 \cr & \Rightarrow x = 50 - 34 \cr & \Rightarrow x = 16 \cr} $$

$$\eqalign{ & {S_n} = \frac{n}{2}\left( {a + l} \right) \cr & {S_9} = \frac{9}{2}\left( {7 + 55} \right) \cr & {S_9} = \frac{9}{2} \times 62 \cr & {S_9} = 9 \times 31 \cr & \,\,\,\,\,\,\,\,\, = 279 \cr} $$

Let the first term of the AP be a and the common difference = d
7^th term = A₇ = a + 6d
11^th term = A₁₁ = a + 10d
According to question,
⇒ 7 × (a + 6d) = 11 × (a + 10d)
⇒ 7a + 42d = 11a + 110d
⇒ 11a - 7a = 42d - 110d
⇒ 4a = -68d
⇒ a = -17d
⇒ a + 17d = 0 = A₁₈

T₇ = a + 6d = -15 ....... (1)
T₁₂ = a + 11d = 5 ...... (2)
on solving (1) and (2)
d = 4 & a = -39
T₁₆ = a + 15d
      = -39 + 15(4)
      = -39 + 60
      = 21