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Ex 9.2
Ex 9.2, 2 Deleted for CBSE Board 2023 Exams
Ex 9.2, 3 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 4 Deleted for CBSE Board 2023 Exams
Ex 9.2, 5 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 6 Deleted for CBSE Board 2023 Exams
Ex 9.2, 7 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 8 Deleted for CBSE Board 2023 Exams
Ex 9.2, 9 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 10 Deleted for CBSE Board 2023 Exams
Ex 9.2, 11 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 12 Deleted for CBSE Board 2023 Exams
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Ex 9.2, 14 Important
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Ex 9.2
Last updated at May 29, 2018 by Teachoo
Ex 9.2,13 If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m. Let a1, a2, … an be the given A.P Given, Sum of n terms = 3n2 + 5n Sn = 3n2 + 5n Putting n = 1 in (1) S1 = 3 × 12 + 5 × 1 = 3 × 1 + 5 × 1 = 3 + 5 = 8 Sum of first 1 terms = First term ∴ First term = a1 = S1 = 8 Sn = 3n2 + 5n …(1) Putting n = 2 in (1) S2 = 3 × 22 + 5 × 2 = 3 × 4 + 5 × 2 = 12 + 10 = 22 Sum of first two terms = First term + Second term S2 = a1 + a2 S2 – a1 = a2 a2 = S2 – a1 Putting a1 = 8 , S2 = 22 a2 = 22 – 8 = 14 Thus, a1 = 8 , a2 = 14 Common difference (d) = Second term – First term = a2 – a1 = 14 – 8 = 6 Now we have a = 8 , d = 6 We know that, an = a + (n – 1)d where an = nth term of A.P. n = number of terms a = first term , d = common difference Given that mth term = 164 Putting n = m in an mth term = am = a + (m – 1)d Putting a = 8, d= 6 & mth term = 164 164 = 8 + (m – 1)6 164 – 8 = (m – 1)6 156 = (m – 1)6 156/6 = m – 1 26 = m – 1 26 + 1 = m 27 = m m = 27 Hence, Value of m is 27