math book class 10 exercise 5.3 ka question no. 3
Answers
Step-by-step explanation:In an AP(i) Given a = 5, d = 3, an = 50, find n and Sn.(ii) Given a = 7, a13 = 35, find d and S13.(iii) Given a12 = 37, d = 3, find a and S12.(iv) Given a3 = 15, S10 = 125, find d and a10.(v) Given d = 5, S9 = 75, find a and a9.(vi) Given a = 2, d = 8, Sn = 90, find n and an.(vii) Given a = 8, an = 62, Sn = 210, find n and d.(viii) Given an = 4, d = 2, Sn = − 14, find n and a.(ix) Given a = 3, n = 8, S = 192, find d.(x) Given l = 28, S = 144 and there are total 9 terms. Find a.
Solutions:
(i) Given that, a = 5, d = 3, an = 50As we know, from the formula of the nth term in an AP,an = a +(n −1)d,Therefore, putting the given values, we get,⇒ 50 = 5+(n -1)×3⇒ 3(n -1) = 45⇒ n -1 = 15⇒ n = 16Now, sum of n terms,Sn = n/2 (a +an)Sn = 16/2 (5 + 50) = 440
(ii) Given that, a = 7, a13 = 35As we know, from the formula of the nth term in an AP,an = a+(n−1)d,Therefore, putting the given values, we get,⇒ 35 = 7+(13-1)d⇒ 12d = 28⇒ d = 28/12 = 2.33Now, Sn = n/2 (a+an)S13 = 13/2 (7+35) = 273
(iii)Given that, a12 = 37, d = 3As we know, from the formula of the nth term in an AP,an = a+(n −1)d,Therefore, putting the given values, we get,⇒ a12 = a+(12−1)3⇒ 37 = a+33⇒ a = 4Now, sum of nth term,Sn = n/2 (a+an)Sn = 12/2 (4+37)= 246
(iv) Given that, a3 = 15, S10 = 125As we know, from the formula of the nth term in an AP,an = a +(n−1)d,
Therefore, putting the given values, we get,a3 = a+(3−1)d15 = a+2d ………………………….. (i)Sum of the nth term,Sn = n/2 [2a+(n-1)d]S10 = 10/2 [2a+(10-1)d]125 = 5(2a+9d)25 = 2a+9d ……………………….. (ii)On multiplying equation (i) by (ii), we will get;30 = 2a+4d ………………………………. (iii)By subtracting equation (iii) from (ii), we get,−5 = 5dd = −1From equation (i),15 = a+2(−1)15 = a−2a = 17 = First terma10 = a+(10−1)da10 = 17+(9)(−1)a10 = 17−9 = 8(v) Given that, d = 5, S9 = 75As, sum of n terms in AP is,Sn = n/2 [2a +(n -1)d]Therefore, the sum of first nine terms are;S9 = 9/2 [2a +(9-1)5]25 = 3(a+20)25 = 3a+603a = 25−60a = -35/3As we know, the nth term can be written as;an = a+(n−1)da9 = a+(9−1)(5)= -35/3+8(5)= -35/3+40= (35+120/3) = 85/3(vi) Given that, a = 2, d = 8, Sn = 90As, sum of n terms in an AP is,Sn = n/2 [2a +(n -1)d]90 = n/2 [2a +(n -1)d]⇒ 180 = n(4+8n -8) = n(8n-4) = 8n2-4n⇒ 8n2-4n –180 = 0⇒ 2n2–n-45 = 0⇒ 2n2-10n+9n-45 = 0⇒ 2n(n -5)+9(n -5) = 0⇒ (2n-9)(2n+9) = 0So, n = 5 (as it is positive integer)∴ a5 = 8+5×4 = 34(vii) Given that, a = 8, an = 62, Sn = 210As, sum of n terms in an AP is,Sn = n/2 (a + an)210 = n/2 (8 +62)⇒ 35n = 210⇒ n = 210/35 = 6Now, 62 = 8+5d⇒ 5d = 62-8 = 54⇒ d = 54/5 = 10.8(viii) Given that, nth term, an = 4, common difference, d = 2, sum of n terms, Sn = −14.As we know, from the formula of the nth term in an AP,an = a+(n −1)d,Therefore, putting the given values, we get,4 = a+(n −1)24 = a+2n−2a+2n = 6a = 6 − 2n …………………………………………. (i)As we know, the sum of n terms is;Sn = n/2 (a+an)-14 = n/2 (a+4)−28 = n (a+4)−28 = n (6 −2n +4) {From equation (i)}−28 = n (− 2n +10)−28 = − 2n2+10n2n2 −10n − 28 = 0n2 −5n −14 = 0n2 −7n+2n −14 = 0n (n−7)+2(n −7) = 0(n −7)(n +2) = 0Either n − 7 = 0 or n + 2 = 0n = 7 or n = −2However, n can neither be negative nor fractional.
Therefore, n = 7
From equation (i), we geta = 6−2na = 6−2(7)= 6−14= −8(ix) Given that, first term, a = 3,Number of terms, n = 8And sum of n terms, S = 192As we know,Sn = n/2 [2a+(n -1)d]192 = 8/2 [2×3+(8 -1)d]192 = 4[6 +7d]48 = 6+7d42 = 7dd = 6(x) Given that, l = 28,S = 144 and there are total of 9 terms.Sum of n terms formula,Sn = n/2 (a + l)144 = 9/2(a+28)(16)×(2) = a+2832 = a+28a = 4
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