Question

if the 10th term of an ap is 52 and 17th term is 20 more than the 13th term find the AP class 10 ​

Answer 1

◘ Given ◘

The 10th term of an AP is 52 and 17th term is 20 more than the 13th term.

• a₁₀ = 52
• a₁₇ = 20 + a₁₃

◘ To Find ◘

The arithmetic progression (AP) whose certain terms are given.

◘ Solution ◘

We know,

$\boxed{\bf{a_n=a+(n-1)d}}$

We get :

a₁₀ = a + (10 - 1)d = 52

→ a + 9d = 52           . . . (i)

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A/q,

a₁₇ = 20 + a₁₃

→ a + (17 - 1)d = 20 + [ a + (13 - 1)d ]

→ a + 16d = 20 + a + 12d

→ 16d - 12d + a - a = 20

→ 4d = 20

→ d = 20/4

→ d = 5

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◙ Putting the value of d in equation (i) :-

a + 9d = 52

→ a + 9(5) = 52

→ a + 45 = 52

→ a = 52 - 45

→ a = 7

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Now, the AP is :-

• a₁ = 7

• a₂ = a + d = 7 + 5 = 12

• a₃ = a + 2d = 7 + 2(5) = 17

• a₄ = a + 3d = 7 + 3(5) = 22

Therefore, the AP is 7, 12, 17, 22, 27, ....

Answer 2

GIVEN :–

• 10th term of A.P. is 52.

• And 17th term is 20 more than the 13th term.

TO FIND :–

• A.P. = ?

SOLUTION :–

• We know that –

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$\bf \implies \large{ \boxed{ \bf T_n = a+(n-1)d}}$

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• According to the first condition –

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$\bf \implies T_{10}= 52$

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$\bf \implies a + (10 - 1)d= 52$

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$\bf \implies a +9d= 52 \: \: \: \: - - - eq.(1)$

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• According to the second condition –

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$\bf \implies T_{17}=T_{13} + 20$

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$\bf \implies a + (17 - 1)d=a + (13 - 1)d+ 20$

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$\bf \implies a +16d=a +12d+ 20$

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$\bf \implies 16d=12d+ 20$

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$\bf \implies 16d - 12d = 20$

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$\bf \implies 4d = 20$

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$\bf \implies d = \dfrac{20}{4}$

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$\bf \implies \large{ \boxed{ \bf d=5}}$

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• Now put the value of 'd' in eq.(1) –

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$\bf \implies a +9(5)= 52$

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$\bf \implies a +45= 52$

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$\bf \implies a = 52 - 45$

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$\bf \implies \large{ \boxed{ \bf a=5}}$

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• Hence –

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$\bf \implies A.P. = 7,7 + 5,7 + (2)5,........$

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$\bf \implies A.P. = 7,12,17,........$