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(i) 2, 4, 8, 16 … (ii) 2, 5/2, 3, 7/2 …. (iii) -1.2, -3.2, -5.2, -7.2 … (iv) -10, – 6, – 2, 2 … (v) 3, 3 + √2, 3 + 2√2, 3 + 3√2 (vi) 0.2, 0.22, 0.222, 0.2222 …. (vii) 0, – 4, – 8, – 12 … (viii) -1/2, -1/2, -1/2, -1/2 …. (ix) 1, 3, 9, 27 … (x) a, 2a, 3a, 4a … (xi) a, a2, a3, a4 … (xii) √2, √8, √18, √32 … (xiii) √3, √6, √9, √12 … (xiv) 12, 32, 52, 72 … (xv) 12, 52, 72, 73 …

(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ….
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, – 6, – 2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0, – 4, – 8, – 12 …
(viii) -1/2, -1/2, -1/2, -1/2 ….
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a …
(xi) a, a2, a3, a4 …
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 12, 32, 52, 72 …
(xv) 12, 52, 72, 73 …

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 8740 Points
4 months ago
(i) Given to us, 2, 4, 8, 16 … Here, the common difference is; a2 – a1 = 4 – 2 = 2 a3 – a2 = 8 – 4 = 4 a4 – a3 = 16 – 8 = 8 Since, an+1 – an or the common difference is not the same every time. Therefore, the given series are not forming an A.P. (ii) Given, 2, 5/2, 3, 7/2 …. Here, a2 – a1 = 5/2-2 = 1/2 a3 – a2 = 3-5/2 = 1/2 a4 – a3 = 7/2-3 = 1/2 Since, an+1 – an or the common difference is same every time. Therefore, d = 1/2 and the given series are in A.P. The next three terms are; a5 = 7/2+1/2 = 4 a6 = 4 +1/2 = 9/2 a7 = 9/2 +1/2 = 5 (iii) Given, -1.2, – 3.2, -5.2, -7.2 … Here, a2 – a1 = (-3.2)-(-1.2) = -2 a3 – a2 = (-5.2)-(-3.2) = -2 a4 – a3 = (-7.2)-(-5.2) = -2 Since, an+1 – an or common difference is same every time. Therefore, d = -2 and the given series are in A.P. Hence, next three terms are; a5 = – 7.2-2 = -9.2 a6 = – 9.2-2 = – 11.2 a7 = – 11.2-2 = – 13.2 (iv) Given, -10, – 6, – 2, 2 … Here, the terms and their difference are; a2 – a1 = (-6)-(-10) = 4 a3 – a2 = (-2)-(-6) = 4 a4 – a3 = (2 -(-2) = 4 Since, an+1 – an or the common difference is same every time. Therefore, d = 4 and the given numbers are in A.P. Hence, next three terms are; a5 = 2+4 = 6 a6 = 6+4 = 10 a7 = 10+4 = 14 (v) Given, 3, 3+√2, 3+2√2, 3+3√2 Here, a2 – a1 = 3+√2-3 = √2 a3 – a2 = (3+2√2)-(3+√2) = √2 a4 – a3 = (3+3√2) – (3+2√2) = √2 Since, an+1 – an or the common difference is same every time. Therefore, d = √2 and the given series forms a A.P. Hence, next three terms are; a5 = (3+√2) +√2 = 3+4√2 a6 = (3+4√2)+√2 = 3+5√2 a7 = (3+5√2)+√2 = 3+6√2 (vi) 0.2, 0.22, 0.222, 0.2222 …. Here, a2 – a1 = 0.22-0.2 = 0.02 a3 – a2 = 0.222-0.22 = 0.002 a4 – a3 = 0.2222-0.222 = 0.0002 Since, an+1 – an or the common difference is not same every time. Therefore, and the given series doesn’t forms a A.P. (vii) 0, -4, -8, -12 … Here, a2 – a1 = (-4)-0 = -4 a3 – a2 = (-8)-(-4) = -4 a4 – a3 = (-12)-(-8) = -4 Since, an+1 – an or the common difference is same every time. Therefore, d = -4 and the given series forms a A.P. Hence, next three terms are; a5 = -12-4 = -16 a6 = -16-4 = -20 a7 = -20-4 = -24 (viii) -1/2, -1/2, -1/2, -1/2 …. Here, a2 – a1 = (-1/2) – (-1/2) = 0 a3 – a2 = (-1/2) – (-1/2) = 0 a4 – a3 = (-1/2) – (-1/2) = 0 Since, an+1 – an or the common difference is same every time. Therefore, d = 0 and the given series forms a A.P. Hence, next three terms are; a5 = (-1/2)-0 = -1/2 a6 = (-1/2)-0 = -1/2 a7 = (-1/2)-0 = -1/2 (ix) 1, 3, 9, 27 … Here, a2 – a1 = 3-1 = 2 a3 – a2 = 9-3 = 6 a4 – a3 = 27-9 = 18 Since, an+1 – an or the common difference is not same every time. Therefore, and the given series doesn’t form a A.P. (x) a, 2a, 3a, 4a … Here, a2 – a1 = 2a–a = a a3 – a2 = 3a-2a = a a4 – a3 = 4a-3a = a Since, an+1 – an or the common difference is same every time. Therefore, d = a and the given series forms a A.P. Hence, next three terms are; a5 = 4a+a = 5a a6 = 5a+a = 6a a7 = 6a+a = 7a (xi) a, a2, a3, a4 … Here, a2 – a1 = a2–a = a(a-1) a3 – a2 = a3 – a2 = a2(a-1) a4 – a3 = a4 – a3 = a3(a-1) Since, an+1 – an or the common difference is not same every time. Therefore, the given series doesn’t forms a A.P. (xii) √2, √8, √18, √32 … Here, a2 – a1 = √8-√2 = 2√2-√2 = √2 a3 – a2 = √18-√8 = 3√2-2√2 = √2 a4 – a3 = 4√2-3√2 = √2 Since, an+1 – an or the common difference is same every time. Therefore, d = √2 and the given series forms a A.P. Hence, next three terms are; a5 = √32+√2 = 4√2+√2 = 5√2 = √50 a6 = 5√2+√2 = 6√2 = √72 a7 = 6√2+√2 = 7√2 = √98 (xiii) √3, √6, √9, √12 … Here, a2 – a1 = √6-√3 = √3×√2-√3 = √3(√2-1) a3 – a2 = √9-√6 = 3-√6 = √3(√3-√2) a4 – a3 = √12 – √9 = 2√3 – √3×√3 = √3(2-√3) Since, an+1 – an or the common difference is not same every time. Therefore, the given series doesn’t form a A.P. (xiv) 12, 32, 52, 72 … Or, 1, 9, 25, 49 ….. Here, a2 − a1 = 9−1 = 8 a3 − a2 = 25−9 = 16 a4 − a3 = 49−25 = 24 Since, an+1 – an or the common difference is not same every time. Therefore, the given series doesn’t form a A.P. (xv) 12, 52, 72, 73 … Or 1, 25, 49, 73 … Here, a2 − a1 = 25−1 = 24 a3 − a2 = 49−25 = 24 a4 − a3 = 73−49 = 24 Since, an+1 – an or the common difference is same every time. Therefore, d = 24 and the given series forms a A.P. Hence, next three terms are; a5 = 73+24 = 97 a6 = 97+24 = 121 a7 = 121+24 = 145

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