(i) Given that, a = 5, d = 3, an = 50 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, ⇒ 50 = 5+(n -1)×3 ⇒ 3(n -1) = 45 ⇒ n -1 = 15 ⇒ n = 16 Now, sum of n terms, Sn = n/2 (a +an) Sn = 16/2 (5 + 50) = 440 (ii) Given that, a = 7, a13 = 35 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, ⇒ 35 = 7+(13-1)d ⇒ 12d = 28 ⇒ d = 28/12 = 2.33 Now, Sn = n/2 (a+an) S13 = 13/2 (7+35) = 273 (iii) Given that, a12 = 37, d = 3 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, ⇒ a12 = a+(12−1)3 ⇒ 37 = a+33 ⇒ a = 4 Now, sum of nth term, Sn = n/2 (a+an) Sn = 12/2 (4+37) = 246 (iv) Given that, a3 = 15, S10 = 125 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, a3 = a+(3−1)d 15 = 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 = 5d d = −1 From equation (i), 15 = a+2(−1) 15 = a−2 a = 17 = First term a10 = a+(10−1)d a10 = 17+(9)(−1) a10 = 17−9 = 8 (v) Given that, d = 5, S9 = 75 As, 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+60 3a = 25−60 a = -35/3 As we know, the nth term can be written as; an = a+(n−1)d a9 = a+(9−1)(5) = -35/3+8(5) = -35/3+40 = (35+120/3) = 85/3 (vi) Given that, a = 2, d = 8, Sn = 90 As, 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 ⇒ (n-5)(2n+9) = 0 So, n = 5 (as n only be a positive integer) ∴ a5 = 8+5×4 = 34 (vii) Given that, a = 8, an = 62, Sn = 210 As, sum of n terms in an AP is, Sn = n/2 (a + an) 210 = n/2 (8 +62) ⇒ 35n = 210 ⇒ n = 210/35 = 6 Now, 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)2 4 = a+2n−2 a+2n = 6 a = 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+10n 2n2 −10n − 28 = 0 n2 −5n −14 = 0 n2 −7n+2n −14 = 0 n (n−7)+2(n −7) = 0 (n −7)(n +2) = 0 Either n − 7 = 0 or n + 2 = 0 n = 7 or n = −2 However, n can neither be negative nor fractional. Therefore, n = 7 From equation (i), we get a = 6−2n a = 6−2(7) = 6−14 = −8 (ix) Given that, first term, a = 3, Number of terms, n = 8 And sum of n terms, S = 192 As we know, Sn = n/2 [2a+(n -1)d] 192 = 8/2 [2×3+(8 -1)d] 192 = 4[6 +7d] 48 = 6+7d 42 = 7d d = 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+28 32 = a+28 a = 4