Question

find the 20th term of an ap whose 3rd term is 7 and the seventh term exceed three times the 3rd term by 2 also find it​

Answer 1

EXPLANATION.

3rd term of an Ap = 7. .........(1)

seventh term = 3 ( 3rd term) / 2.

To find the 20th term of an Ap.

Nth term of an Ap.

An = a + ( n - 1 )d.

→ a + 2d = 7 ........... (1).

→ a + 6d = 3 ( a + 2d ) / 2.

→ 2 ( a + 6d ) = 3 ( a + 2d ).

→ 2a + 12d = 3a + 6d.

→ 12d - 6d = 3a - 2a.

→ 6d = a .......... (2).

put the value of equation (2) in (1).

we get,

→ 6d + 2d = 7.

→ 8d = 7.

→ d = 7/8.

put the value of D = 7/8 in equation (2)

we get,

→ 6 X 7/8 = a.

→ 21/4 = a.

→ First term = a = 21/4.

→ Common difference = 7/8.

20th term of an Ap.

→ a + 19d.

→ 21/4 + 19 X 7/8.

→ 21/4 + 133/8.

→ 42 + 133 / 8.

→ 175/8.

20th term of an Ap = 175/8.

Answer 2

Given :-

• $\sf a_3 = 7$
• $\sf a_7 = \frac{3 \times a_3}{2}$

To Find :-

• $\sf a_{20}$

Solution :-

We know that, For nth term of A.P. -

$\sf a_n = a + (n-1)d$

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So, for 3rd term -

➩ $\sf a_3 = a + ( 3 - 1 )d$

➩ $\sf a_3 = a + 2d$

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We are given that 3rd term is 7, So -

➩ $\sf a + 2d = 7 \: \: \: \: \: \: -i$

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For 7th term -

➩ $\sf a_7 = a + (7-1)d$

➩ $\sf a_7 = a + 6d$

We are given that - $\sf a_7 = \frac{3 \times a_3}{2}$

➩ $\sf a + 6d = \frac{3 \times a_3}{2}$

➩ $\sf a + 6d = \frac{3 \times 7}{2}$

➩ $\sf a + 6d = 10.5 \: \: \: \: \: \: -ii$

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Subtracting equation i from ii

➩ $\sf a + 6d - a - 2d = 10.5 - 7$

➩ $\sf 4d = 3.5$

➩ $\sf d = \frac{3.5}{4}$

➩ $\sf d = 0.875$

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Substituting the value of d in equation i

➩ $\sf a + 2d = 7$

➩ $\sf a + 2 \times 0.875 = 7$

➩ $\sf a = 7 - 1.75$

➩ $\sf a = 5.25$

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• Common difference (d) = 0.875
• First term (a) = 5.25

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Calculating 20th term -

➩ $\sf a_{20} = a + (20-1)d$

➩ $\sf a_{20} = 5.25 + 19 \times 0.875$

➩ $\sf a_{20} = 5.25 + 16.625$

➩ $\sf a_{20} = 21.875$

20th term of A.P. = 21.875