Question

the 3th term of an a.p is 1 and the 6th term is -11.find its 15th term.​

Answer 1

A n s w e r

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G i v e n

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• 3rd term of an A.P is 1

• 6th term of that A.P is -11

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F i n d

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• 15th term of the A.P

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S o l u t i o n

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We know that ,

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➠ $\sf a_n = a + (n - 1)d$ ⚊⚊⚊⚊ ⓵

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Where ,

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• $\sf a_n$ = nth term

• a = First term

• n = Number of terms

• d = Common difference

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$\underline{\bold{\texttt{For 3rd term :}}}$

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• $\sf a_n$ = 1

• a = a

• n = 3

• d = d

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⟮ Putting the above values in ⓵ ⟯

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➜ $\sf a_n = a + (n - 1)d$

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➜ $\sf 1 = a + (3 - 1)d$

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➜ 1 = a + 2d ⚊⚊⚊⚊ ⓶

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$\underline{\bold{\texttt{For 6th term :}}}$

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• $\sf a_n$ = -11

• a = a

• n = 6

• d = d

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⟮ Putting the above values in ⓵ ⟯

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➜ $\sf a_n = a + (n - 1)d$

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➜ $\sf -11 = a + (6 - 1)d$

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➜ -11 = a + 5d ⚊⚊⚊⚊ ⓷

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⟮ Equation ⓷ - ⓶ ⟯

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➜ -11 - 1 = a + 5d - (a + 2d)

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➜ -12 = a + 5d - a - 2d

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➜ -12 = 3d

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➨ d = -4 ⚊⚊⚊⚊ ⓸

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• Hence the common difference is -4

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⟮ Putting d = -4 from ⓸ to ⓶ ⟯

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➜ 1 = a + 2d

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➜ 1 = a + 2(-4)

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➜ 1 = a - 8

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➜ a = 8 + 1

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➨ a = 9 ⚊⚊⚊⚊ ⓹

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• Hence the first term is 9

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$\underline{\bold{\texttt{For 15th term :}}}$

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• $\sf a_n = a_{15}$

• a = 9

• n = 15

• d = -4

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⟮ Putting the above values in ⓵ ⟯

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➜ $\sf a_n = a + (n - 1)d$

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➜ $\sf a_{15} = 9 + (15 - 1)-4$

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➜ $\sf a_{15} = 9 + 14(-4)$

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➜ $\sf a_{15} = 9 - 56$

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➨ $\sf a_{15} = -47$

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• Hence the 15th term of the given A.P is -47