Question

In an AP given third term is 15 and sum of first 15 terms is 125 find 10th term

Answer 1

GIVEN :–

• Third term = 15

• Sum of first 15 terms = 125

TO FIND :–

• 10 th term = ?

SOLUTION :–

• If first term is 'a' & Common difference is 'd'. Then nth term –

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

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• According to the question –

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$\implies \bf T_{3} = 15$

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$\implies \bf a + (3 - 1)d = 15$

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$\bf \implies a + 2d = 15 \: \: \: \: - - - eq.(1)$

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• We also know that Sum of n - terms is –

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$\bf \large \implies { \boxed{ \bf S_{n} = \dfrac{n}{2} \{2a + (n - 1)d \}}}$

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• According to the question –

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$\bf \implies S_{15} = 125$

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$\bf \implies \dfrac{15}{2} \{2a + (15- 1)d \} = 125$

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$\bf \implies \dfrac{15}{2} (2a + 14d) = 125$

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$\bf \implies 15 (a + 7d) = 125$

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$\bf \implies a + 7d = \dfrac{25}{3}$

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• Using eq.(1) –

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$\bf \implies 15 - 2d + 7d = \dfrac{25}{3}$

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$\bf \implies 15 + 5d = \dfrac{25}{3}$

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$\bf \implies 5d = \dfrac{25}{3} - 15$

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$\bf \implies 5d = \dfrac{25 - 45}{3}$

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$\bf \implies 5d = \dfrac{ - 20}{3}$

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$\bf \implies \large{ \boxed{ \bf d = \dfrac{-4}{3}}}$

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• Again Using eq.(1) –

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$\bf \implies a + 2 \left( \dfrac{ - 4}{3} \right) = 15$

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$\bf \implies a - \dfrac{ 8}{3} = 15$

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$\bf \implies a= 15 + \dfrac{ 8}{3}$

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$\bf \implies a= \dfrac{45 + 8}{3}$

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$\bf \implies \large{ \boxed{ \bf a= \dfrac{53}{3}}}$

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• Now Let's find 10th term –

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$\bf \implies T_{10} =\dfrac{53}{3}+ (10 - 1) \left(\dfrac{-4}{3} \right)$

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$\bf \implies T_{10} =\dfrac{53}{3}+9 \left(\dfrac{-4}{3} \right)$

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$\bf \implies T_{10} =\dfrac{53}{3} - \dfrac{36}{3}$

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$\bf \implies T_{10} =\dfrac{53 - 36}{3}$

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$\bf \implies \large{ \boxed{ \bf T_{10} =\dfrac{17}{3}}}$