'Kolam' is a decorative art which is made with rice flour in South Indian States. It is drawn on grid pattern of dots.…

CBSE Class 10 Maths PYQ · Arithmetic Progressions · Word Problem & Applications · 4 Marks · March 2026 · Standard

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1184 Marks · March 2026 · Standard
'Kolam' is a decorative art which is made with rice flour in South Indian States. It is drawn on grid pattern of dots. One such art work is shown below.
Observe the given figure carefully. There are $4$ dots in first square, $8$ dots in second square, $12$ dots in third square and so on.
Based on the above, answer the following questions :
(i) Show that number of dots given above form an A.P. Write the first term and common difference.
(ii) Write $n^{th}$ term of the A.P. formed.
(iii) (a) The pattern is expanded on a large ground. If total $220$ dots are used, then find the number of squares formed.
OR
(b) Is it possible to complete $n$ number of squares using $100$ dots ? If yes, then find the value of $n$.
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Sol. (i) Number of dots formed in each square are $4, 8, 12, ...$
$12 - 8 = 8 - 4 = 4$
$\therefore$ Numbers of dots form an A.P. (I) ($\frac{1}{2}$ Mark)
Here, $a = 4, d = 4$ (II) ($\frac{1}{2}$ Mark)
(ii) $a_n = 4 + (n - 1) 4 = 4n$ (I) (1 Mark)
(iii) (a) Here, $a = 4, d = 4, S_n = 220$
$\therefore 220 = \frac{n}{2} (2 \times 4 + (n - 1)4)$ (I) ($\frac{1}{2}$ Mark)
$\Rightarrow (n + 1) n = 110$
$\Rightarrow n^2 + n - 110 = 0$
$\Rightarrow (n + 11) (n - 10) = 0$
$\Rightarrow n = -11, 10$
$n \neq -11$
$\Rightarrow n = 10$ (II) ($\frac{1}{2}$ Mark)
$\therefore$ The number of squares formed $= 10$
OR
(b) Here $S_n = 100, a = 4, d = 4$
$\therefore 100 = \frac{n}{2} (2 \times 4 + (n - 1)4)$ (I) ($\frac{1}{2}$ Mark)
$\Rightarrow 100 = \frac{4n}{2} (n + 1)$
$\Rightarrow n (n + 1) = 50$
$\Rightarrow n^2 + n - 50 = 0$
$\Rightarrow n = \frac{-1 \pm \sqrt{1+200}}{2} = \frac{-1 \pm \sqrt{201}}{2}$ (II) ($\frac{1}{2}$ Mark)
$\therefore n$ is not a natural number.
$\therefore$ It is not possible to draw complete squares using $100$ dots. (III) (1 Mark)
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