Question

2. While playing a treasure hunt game, some clues are hidden in various spots collectively forms

an AP. If the number on the nth spot is 20 + 4, then answer the following questions to help

the player in spotting the clues.


(a) Which number is on the first spot?

(a) 20 (b) 24 (c) 16 (d) 28

(b) Which number is on the ( − 2)

ℎ

spot?

(a) 16+4n (b) 24+4n (c) 12+4n (d) 28+4n


(c) Which number is on the 34th spot?

(a) 156 (b) 116 (c) 120 (d) 160


(d) What is the sum of all the numbers on the first 10 spots?

(a) 410 (b) 420 (c) 480 (d) 490


(e) Which spot is numbered as 116?

(a) 5th

(b) 8th

(c) 9th

(d) 24th​

Answer 1

The answers are :

(a) The number on the first spot is 24 (option b)

(b) The number on the $(n-2)^{th}$ spot is 12+4n (option c)

(c) The number on $34^{th}$ spot is 156 (option a)

(d) Sum of all numbers on first 10 spots will be 420

(e) The spot which is numbered as 116 is $24^{th}$ spot

Step-by-step explanation:

Given that $n^{th} spot (T_{n} ) = 20 + 4n$ -----(i)

(a) To find the number on the first spot

First spot means that n =1.

Substituting the value of n in (i), we get

T₁ = 20 + 4(1)

T₁  = 20 + 4

T₁  = 24

(b) To find the number on  $(n-2)^{th}$ spot

Here n=n-2

Substituting the value of n in (i)

$T_{n-2} = 20+4(n-2)\\\\T_{n-2} = 20+4n-8\\\\T_{n-2}= 12+4n$

(c) To find the number on $34^{th}$ spot

$34^{th}$ spot means n=34

Substituting the value of n iin equation (i)

T₃₄ = 20+ 4(34)

T₃₄ = 20 +136

T₃₄ = 156

(d) To find the sum of the numbers on the first 10 spots

We know that the sum of an AP is given by the formula,

$S_{n} = \frac{n}{2}(2a+(n-1)d)$ ------ (ii)

where, n = number of terms whose sum is to be calculated , a = First term , d = Common difference

Here we have n = 10 , a = 24

We have to find d (common difference) which can be calculated by T₂ - T₁

T₂ = 20+ 4(2)

T₂= 20+8

T₂ = 28

∴ d = 28 - 24

  d = 4

Now substituting the value of a, n and d in equation (ii)

$S_{10} = \frac{10}{2}(2(24)+(10-1)4)$

$S_{10} = 5(48+(9)4)$

$S_{10} = 5(48+36)$

$S_{10} = 5(84)$

$S_{10} = 420$

(e) To find the spot which is numbered as 116

∴ 116 = 20 + 4n

116-20 = 4n

96 = 4n

24 = n

∴ 116 lies on $24^{th}$ spot