Pawan Prajapati
Last Activity: 3 Years 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
