![Sum of Geometric Series](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhTBWdRmiWVQvJg9sEo-n4OpBBJesx7aOYKeCePCRU_A1oUgEhMcjGiYFwoUHjvmUyw99B-eRuCKKihtsg0RwZhGIzRe0UjZp_fszAj-L_VjL_U4Iu71aLtzcrGt3HkNGkO5kqyyH6UxumPwEGQBK_jCdcc0eI0eM5eyFe0llYdKdKASF7iVaK374R75g/s16000/Sum%20of%20Geometric%20Series.png)
The series associated with a geometric sequence is known as a
geometric series. For example, 2 + 4 + 8 + 16 + 32 is the geometric series
associated with the geometric sequence 2, 4, 8, 16, 32.
Let a be the first term, r be the common ratio, n be the number
of terms, l be the last term and Sn be the sum to n terms of GS, then
Sn = a + ar + ar2 + … … … + arn-2
+ arn-1 ………. (i)
Multiplying both sides by r,
rSn = ar + ar2 + ar3 + … … … +
arn-1 + arn …….. (ii)
Subtracting (i) from (ii), we get
(r – 1)Sn = -a + arn
![∴ Sn = a(r^n-1)/(r-1) Again, Sn = (ar^n-a)/(r-1) or, Sn = (ar^(n-1).r-a)/(r-1) or, Sn = (lr-a)/(r-1) [∵ l = arn-1] ∴ Sn = (lr-a)/(r-1)](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjSeV9lkD_KonLBU8mWPs86zkCeaxZNI3i7i1D3ktYamGYtyv5Q2pa3O44XHKU3PiBitcjFGo02SthpPnM6yQ2lNgHTd_Erjrea6FFB5p1yCVptmvdS2q1s950Afygt0ItTppic3BGpmH9p1uspXLTomDHNULFV8_cYaeVLSl9u7xNNcnO4j3pZlrbAjw/s16000/formula%20Sn%20for%20GS.png)
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Worked Out Example
Example 1: Find the sum of the geometric series 1 + 2 + 4 + … … … 7 terms.
Solution:
Here,
First term (a) = 1
Common ratio (r) = 2/1 = 2
Number of terms (n) = 7
Sum of the terms (Sn) = ?
By using formula,
![Sn = a(r^n-1)/(r-1) or, Sn = 1(2^7-1)/(2-1) or, Sn = (128-1)/(2-1) or, Sn = 127/1 or, Sn = 127](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiG9PsdPGesVEvoqabz7fgvDpUFyACw8Qe6MmHtqB-fL-cogdcKRkzPMhLpjCN3F6SmfnK0yP_lx_8cYAc5IZuqoxsDAipS20nNqXIncJJt70QfKkaUJSqARiXZLWVrCARzZ8LYQZdXjSz-HgEQMNfBNj5ZZyFYgsqruCep8Ba8_67jALJlPJp1Z0y1oQ/s16000/example%201.png)
∴ Sn = 127
Example 2: In a GP, the first term is 7, the last term is 448 and the sum is
889, find the common ratio.
Solution:
Here,
First term (a) = 7
Last term (l) = 448
Sum of terms (Sn) = 889
Common ratio (r) = ?
By using the formula,
![Sn = (lr-a)/(r-1) or, 889 = (448r-7)/(r-1) or, 889r – 889 = 448r – 7 or, 889r – 448r = 889 – 7 or, 441r = 882 or, r = 882/441 or, r = 2](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjo7L7MG3tFZOUGvwVVVWz-TTSzAcwKwUr4eZ-pLC_kux2EvxPpUeYzP3rB1fhNdCV4k3b7KT-cK9BvLYFklsKZmnq4bSk54opGSUUz_qKokA3hw8lfeyspeemq0V1Zl57UfYN5eLuuMbS3NA_HSBgJlc4TSXlLk-2uinh7kXORhV-DPCf-rOLe5FjChA/s16000/example%202.png)
∴ The common ratio is 2.
![Example 3: Find the sum of ∑_(n=1)^5▒3^(n+1)](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgKdm2CzY3Q0n2XrILtHQhm3x0Tqk8-eltzxT9LWBfAlN7yyvnH8iFLzlg2Ij269-Ajf6dDcXh9l4-fV_ptbQ73U7IBRQR-XRxGZYOlPagmSE563Yuv1PudjVGseFJcd5fB9InXHlBVIkVcnu8XSXDDttGL51sQL4PdSNyKsYTsyZpSus0uofsw84g7MQ/s16000/example%203%20question.png)
Solution:
Here,
![∑_(n=1)^5▒3^(n+1)](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgUnXhQKh_4kGxjcFHskGnAp3q1n5aAe22SLjz2J3-SwRvO6oSwSRdBm6KY3JuOU1jD0uUIhjzXl6N69CmhCZL8SomuC8K70bY33kZNx5mHhBujXqtDIcJ3x1S8XXXJRl5h0elayYISV7OJBgm4eNW6h7FwBM-IkdW12z4fmhPydOZQwrmzBhNrfQ1Djw/s16000/example%203%20solution.png)
= 32 + 33 + 34 + 35
+ 36
= 9 + 27 + 81 + 243 + 729
= 1089
Example 4: How many terms of the series 32 + 48 + 72 + … … will add upto 665?
Solution:
Here,
First term (a) = 32
Common ratio (r) = 48/32 = 3/2
Sum of terms (Sn) = 665
Number of terms (n) = ?
By using formula,
![Sn = a(r^n-1)/(r-1) or, 665 = 32{(3/2)^n-1}/(3/2-1) or, 665 = 32{(3/2)^n-1}/(1/2) or, 665 = 64{(3/2)^n-1} or, 665/64 = (3/2)^n-1 or, 665/64 + 1 = (3/2)^n or, (665+64)/64 = (3/2)^n or, 729/64 = (3/2)^n or, (3/2)^6 = (3/2)^n](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhCPSuUdO06_zg7rGR_zGmtU9XFJYzEfvkg1gwU-f38u20_HFdvPZ-_YX_hefzjZ24RphBpar1wZRqBX6qwlcCL9DZ91UBquCas3zjBo72tX_FN9jxpX-OpzwvdhfKBGuGgnMutjRWUyzwuj08Jk_UqufUyCDo_UDL1RsNpUHYwIiQcLkdncnyv9RfmEQ/s16000/example%204.png)
∴ n = 6
∴ The number of terms = 6
Example 5: If S3 and S6 of a GS are 7 and 63
respectively, find the common ratio.
Solution:
Here,
![S3 = 7 or, a(r^3-1)/(r-1) = 7 ……………. (i) S6 = 63 or, a(r^6-1)/(r-1) = 63 …………. (ii) Dividing equation (ii) by (i), (a(r^6-1)/(r-1))/(a(r^3-1)/(r-1)) = 63/7 or, a(r^3+1)(r^3-1)/(r-1) × (r-1)/a(r^3-1) = 63/7](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjhc3b18NF4lV98PCGfwhNh-yAqISpyQlWICgVGXRlW7S4qz9obHbtCoDO2-aSrqKONYmp2jeHPzZ6q3D1Z-8cGPTL0PeDiEWpJXHvHpho-SlBKtew54TFUjubMzA9dbcJLhUCUoVTk1yNwzzIQKhGw9gR5E68FmDjCH2QkYBuvRcv-_-9nJas0YGBORg/s16000/example%205.png)
or, r3 = 9 – 1
or, r3 = 8
or, r3 = 23
or, r = 2
∴ Common ratio = 2
Example 6: The second and fifth terms of GS are 3 and 81 respectively. Find the
sum of the first five terms.
Solution:
Here,
The second term of GS is 3
i.e. t2 = 3
or, ar = 3
Again, the fifth term of GS is 81
i.e. t5 = 81
or, ar4 = 81
or, ar.r3 = 81
or, 3.r3 = 81
or, r3 = 81/3
or, r3 = 27 = 33
or, r = 3
And, ar = 3
or, a×3 = 3
or, a = 1
Now,
![S5 = a(r^5-1)/(r-1) = 1(3^5-1)/(3-1) = (243-1)/2 = 242/2 = 121.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg4Qr37IzTUIiwJ3MexgzOQAmgtzhixhY9pcLcAo8FKvF0Zls38xXEutq9l9gNivbY6bfJqx5dLHsVH3_PWPnKnTrV9K95WevTmU581GfjVJG2qZ0NL5RruVDpMelETgXSJJ_BJaW9Q2eY2ezCMOXZcvfi2zdp3oLzjJfhPHCIxn7ccbis_qowyU3lmcQ/s16000/example%206.png)
Example 7: Find the GP for which the sum of the first two terms is -4 and the
fifth term is 4 times the third term.
Solution:
Here,
S2 = -4
i.e. t1 + t2 = -4
or, a + ar = -4
or, a(1 + r) = -4 ……………… (i)
t5 = 4 × t3
i.e. ar4 = 4 × ar2
or, r2 = 4
or, r = ±2
Taking r = 2, from equation (i), we have
a(1 + 2) = -4
or, a = -4/3
The GP is a, ar, ar2, …
i.e. -4/3, -4×2/3, -4×22/3, …
i.e. -4/3, -8/3, -16/3, …
Taking r = -2, from equation (i), we have
a(1 – 2) = -4
or, a = 4
The GP is a, ar, ar2, …
i.e. 4, 4×-2, 4×(-2)2, …
i.e. 4, -8, 16, …
Example 8: If the sum of the first three terms of a GP is 1 and the sum of the
first six terms is 28. Find the sum of the first 9 terms of the series.
Solution:
Here,
![S3 = 1 or, a(r^3-1)/(r-1) = 1 ……………. (i) S6 = 28 or, a(r^6-1)/(r-1) = 28 …………. (ii) Dividing equation (ii) by (i), (a(r^6-1)/(r-1))/(a(r^3-1)/(r-1)) = 28/1 or, a(r^3+1)(r^3-1)/(r-1) × (r-1)/a(r^3-1) = 28](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhKnaL_IYyMWATvwwWCfBVb86v23YOy7Co9sHhhdO2Hx4az-nRBoJcyurYAOgpZ-LtbMm2s8x-tpWp5HSJOcfOh5NpxLT3H9ByRIQxLNd9Rxdi2OUmvERKikHHLKH3OMVqsYk9_sdBh8GKo-LWCXq1r4GFl__MlzdooL9c1pqe9SOtzkGTbtoCrsOG7fQ/s16000/example%208%20part%201.png)
or, r3 = 28 – 1
or, r3 = 27
or, r3 = 33
or, r = 3
![Substituting r = 3 in (i), a(3^3-1)/(3-1) = 1 or, a(27 – 1) = 2 or, a = 1/13 Now, the sum of the first 9 terms, Sn = a(r^n-1)/(r-1) or, S9 = (1/13(3^9-1))/(3-1) or, S9 = (19683-1)/26 or, S9 = 757](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjS6eKcT2hgZWkugfbpXdazIRXT9E0s3cQktmb1osA2RLnvUr8EbtClUAf8EkVveL8B6xpguIPX0DL7ZH_Gqg0094zBqPMUXGSNl37PkKhazj7XXcAFCOdM0pynp4AMIC_bT-VwaODugBKEi9XHCOrZaacSjpC_DDf--AtDrh0gZMsnf6blIwUOk60zuQ/s16000/example%208%20part%202.png)
∴ The sum of first 9 terms is 757.
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