Progressions – Exercise 2.1 – Class X

1. Which if the following form a sequence?

(i) 4, 11, 18, 28….

(ii) 43, 32, 21, 10….

(iii) 27, 19, 40 , 70….

(iv) 7, 21, 63, 189…

Solution:

(i) 4, 11, 18, 28…. – forms a sequence

(ii) 43, 32, 21, 10….
10, 21, 32, 43…. – forms a sequence

(iii) 27, 19, 40 , 70…. – not a sequence

(iv) 7, 21, 63, 189… – forms a sequence

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2. Write the next two terms of the following sequences

(i) 13, 15, 17, __, ___

^{(ii) 2}/₃, ³/₄, ⁴/₅, __, ___

(iii) 1, 0.1, 0.01, ___, __

(iv) 6, 1, 24, ___, ___

Solution:

(i) 13, 15, 17, 19, 21

^{(ii) 2}/₃, ³/₄, ⁴/₅, ⁵/₆, ⁶/₇

(iii) 1, 0.1, 0.01, 0.001, 0.0001

(iv) 6, 12, 24, 48, 96

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3. If T­_n = 5 – 4n, find first three terms

Solution:

For n = 1,

T­₁ = 5 – 4×1 =  5 – 4 = 1

For n = 2

T­₂ = 5 – 4×2 = 5 – 8 = – 3

For n = 3

T­₃ = 5 – 4×3 = 5 – 12 = – 7

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4. If T­_n = 2n² + 5, find (i) T₃ and T­₁₀

Solution:

If  T­_n = 2n² + 5 then T₃ = 2(3)² +  5

= 2(9) + 5

= 18 + 5

= 23

For T₁₀ = 2(10)² +  5

= 2(100) + 5

= 200 + 5

= 205

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5. If T_n = n² – 1 , find (i) T­_n-1 (ii) T­_n+1

Solution:

If T_n = n² – 1 then, T­_n-1 = (n – 1)² – 1

= n² – 2n + 1 – 1

= n² – 2n

If T_n = n² – 1 then, T­_n+1 = (n+1)² – 1

= n² + 2n + 1 – 1

= n² + 2n

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6. If T­_n = n² + 4 and T­_n = 200, find the value of ‘n’

Solution:

Given, T­_n = n² + 4 and T­_n = 200, we have to find the value of n

200 = n²+ 4

200 – 4 = n²

196 = n²

n = 13

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Next exercise –  Progressions – Exercise 2.2 – Class X

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