हिंदी

Show that the progression $$2, 2\sqrt{2}, 4, 4\sqrt{2}$$, ......... is a G.P. Write its (i) first term (ii) common ratio (iii) nth term (iv) 11th term.

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प्रश्न

Show that the progression $$2, 2\sqrt{2}, 4, 4\sqrt{2}$$, ......... is a G.P.

Write its (i) first term (ii) common ratio (iii) nth term (iv) 11th term.

योग
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उत्तर

To show it is a G.P.

$$\frac{2\sqrt{2}}{2} = \sqrt{2}, \frac{4}{2\sqrt{2}} = \frac{4}{2 \times 1.414...} = \sqrt{2}, \frac{4\sqrt{2}}{4} = \sqrt{2}$$

(Check: $$2\sqrt{2} \times \sqrt{2} = 2 \times 2 = 4$$, so the middle ratio is also $$\sqrt{2}$$.)

All the ratios are the same (constant = $$\sqrt2$$), so the given numbers form a G.P.

(i) First term: $$a = 2$$

(ii) Common ratio: $$r = \sqrt{2}$$

(iii) nth term: $$T_n = ar^{n - 1} = 2 \times (\sqrt{2})^{n-1}$$

(iv) 11th term: $$T_{11} = 2 \times (\sqrt{2})^{10}$$

$$= 2 \times \left((\sqrt{2})^2\right)^5$$

$$= 2 \times 2^5$$

$$= 2 \times 32$$

$$= 64$$

So, the 11th term is 64.

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अध्याय 11: Geometric Progression - EXERCISE 11A [पृष्ठ १७७]

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आर.एस. अग्रवाल Mathematics [English] Class 10 ICSE
अध्याय 11 Geometric Progression
EXERCISE 11A | Q 4. | पृष्ठ १७७
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