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Arithmetic Progressions

Short Answer Type

61. Find the sums given below:
34 + 32 + 30 + . . . + 10

129 Views

62. Find the sums given below:
–5 + (–8) + (–11) + . . . + (–230)

154 Views

63.

In an AP
Given a = 5, d = 3, a_n=50, find n and S_n.

217 Views

64.

In an AP
given a = 7, a₁₃=35, find d and S₁₃.

458 Views

65.

In an AP
Given a₁₂ = 37, d = 3, find ‘a’ and S₁₂.

538 Views

66. Given a₃ = 75, S₁₀ = 120, find ‘d’ and a_10.

Here, a₃= 15
S₁₀ = 120
We know that,

$straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket therefore space space straight S subscript 10 equals 10 over 2 space left square bracket 2 straight a plus left parenthesis 10 minus 1 right parenthesis straight d right square bracket rightwards double arrow space 120 space equals space 5 space straight x space left square bracket 2 straight a plus 9 straight d right square bracket rightwards double arrow 2 straight a plus 9 straight d equals 120 over 5 equals 24 rightwards double arrow space 2 straight a plus 9 straight d space equals space 24 space space space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis And. space space space straight a subscript 1 equals 15 rightwards double arrow space space straight a space plus 2 straight d space equals space 15 space space space space space space space space space space space space space space space... left parenthesis ii right parenthesis$
Solving (i) and (ii)
5d = -6
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$rightwards double arrow space space space straight a space equals space 15 plus 12 over 5 equals fraction numerator 75 plus 12 over denominator 5 end fraction equals 87 over 5 Now comma space space space space straight a subscript 10 equals straight a plus 9 straight d equals 87 over 5 plus 9 open parentheses fraction numerator negative 6 over denominator 5 end fraction close parentheses space space space space space space space space space space space space space space space space space space space space equals 87 over 5 minus 54 over 5 equals fraction numerator 87 minus 54 over denominator 5 end fraction equals 33 over 5 Hence comma space space straight d space equals space fraction numerator negative 6 over denominator 5 end fraction space and space straight a subscript 10 space equals space 33 over 5$

135 Views

67.

In an AP
Given d = 5, 5₉ = 75, find ‘a’ and ‘a’

182 Views

68. Find the sum of the series 5 + 7 + 9+ 10 + 13 + 13 + 17 + 16 +.........to 40 terms.

230 Views

69. Find four numbers in A .P. whose sum is 20 and the sum of whose squares is 180.

253 Views

70. Prove that no matter what the real numbers ‘a’ and ‘b’ are the sequence with nth term a + nb is always an A .P. what is the common difference ? What is the sum of the first 20 terms.

256 Views

Previous Year Papers

Test Series

Test Yourself

Short Answer Type

66. Given a3 = 75, S10 = 120, find ‘d’ and a10.

66. Given a₃ = 75, S₁₀ = 120, find ‘d’ and a_10.