# ICSE Class 10 Geometric Progression Previous Years Questions Solution

This page consists of the solution to previous years questions from the chapter Geometric Progression (G.P.). Geometric Progression is a comparatively new addition to the ICSE Class 10 Maths syllabus. ICSE Class 10 Geometric Progression Previous Years Questions Solution is going to be very beneficial for the students appearing for ICSE class 10 Board exam. The page contains solution to G.P. PYQs which came in ICSE Board exams till the year 2023. You can also download the PDF of the questions from the chapter.

## Download PDF of ICSE Class 10 Geometric Progression Previous Years Questions Solution

1. The 4th, 6th and the last term of a geometric progression are 10, 40 and 640 respectively. If the common ratio is positive, find the first term, common ratio and the number of terms of the series. [2020]

Solution: first term = 1.25, common ratio = 2, no. of terms = 10

Step-by-step Explanation:

$\style{font-size:12px}{4th\;term\;=\;ar^3\;=\;10\;….(i)\\6th\;term\;=\;ar^5\;=\;40\;….(2)\\last\;term\;=\;640\\Dividing\;(2)\;by\;(1),\;we\;get\\\frac{ar^5}{ar^3}=\frac{40}{10}\\r^2\;=\;4\\r=\pm2\\as\;common\;ratio\;is\;positive,\;\\therefore\;r\;=\;2\\Putting\;r=\;2\;in\;(1)\\a\times2^3\;=\;10\\a=\frac{10}8\\a=1.25\\\\let\;nth\;term\;be\;the\;last\;term.\\\therefore ar^{n-1}\;=\;640\\1.25\times2^{n-1}=640\\2^{n-1}=\frac{640}{1.25}\\2^{n-1}=512\\2^{n-1}=2^9\\n-1=9\\n=10}$

ICSE Class 10 Maths PYQ Solution Chapter-wise

Watch video solution of Geometric Progression [G.P.] PYQs here.

2. The first and last trem of a Geometric Progression (G.P.) are 3 and 96 respectively. If the common ratio is 2, find:

(i) ‘n’ the number of terms of the G.P. (ii) sum of n terms [2019]

Solution: (i) 6 (ii) 189

Step-by-step Explanation:

$\style{font-size:12px}{1st\;term\;=\;a\;=\;3\;\\common\;ratio=r=2\\last\;term\;=\;96\\(i)\;Let\;the\;nth\;term\;be\;the\;last\;term.\\\therefore ar^{n-1}=96\\3\times2^{n-1}=96\\2^{n-1}=32\\2^{n-1}=2^5\\n-1=5\\n=6\\\\(ii)\;S_n=\frac{a(r^n-1)}{r-1}\\=\frac{3\;(2^6-1)}{2-1}\\=3\times63\\=189}$

3. The 4th term of a G.P. is 16 and the 7th term is 128. Find the first term and common ratio of the series. [2018]

Solution: first term = 2, common ratio = 2

Step-by-step Explanation:

$\style{font-size:12px}{4th\;term\;=\;ar^3=16\;…(1)\\7th\;term\;=\;ar^6=128\;…(2)\\Dividing\;(2)\;by\;(1),\;we\;get,\\\frac{ar^6}{ar^3}=\frac{128}{16}\\r^3\;=\;8\\r=\sqrt[3]8\\r=2\\Putting\;r=2\;in\;(1)\\a\times2^3=16\\a=\frac{16}8\\a=2}$

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