Question

if 8 th term of an AP exceeds 5th term by 24 and 3rd term of an AP is 17 then find the 15th term of an AP
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Answer 1

Answer

The required 15th term of the AP is 113

$\bf \large \underline{Given : }$

• 8th term of the AP exceeds 5th term by 24
• 3rd term of an AP is 17

$\bf \large \underline{To \: Find : }$

• The 15th term of an AP

$\bf \large \underline{Solution : }$

Let us consider the first term be a and common difference be d

$\sf \underline{ \underline{ \dag \: \: By \: question \: we \: have}}$

$\sf\implies 8th \: term = 5th \: term + 24 \\\\ \sf\implies a + (8-1)d = a + (5-1)d + 24 \\\\ \sf\implies 7d = 4d + 24 \\\\ \sf\implies 3d = 24 \\\\ \sf\implies d = 8$

$\sf \underline{ \underline{ \dag \: \: Again \: we \: have \: from \: question}}$

$\sf\implies 3rd \: term = 17 \\\\ \sf\implies a + (3-1)d = 17 \\\\ \sf\implies a + 2\times 8 = 17 \\\\ \sf\implies a + 16 = 17 \\\\ \sf\implies a = 1$

★$\sf\underline{\underline{Thus \ the \ 15th\ term \ will \ be }}$

$\sf\implies T_{15} = a + (15-1)d \\\\ \sf\implies T_{15} =1 + 14\times 8 \\\\ \sf\implies T_{15} = 1+112 \\\\ \sf\implies T_{15} = 113$

Answer 2

Given ,

• The 8th term of an AP exceeds 5th term by 24

• The 3rd term of AP is 17

Let , the first term and common difference of AP be " a " and " d "

According to the question ,

a + 7d - (a + 4d) = 24

7d - 4d = 24

d = 24/3

d = 8

Since , a + 2d = 17

Thus ,

a + 2(8) = 17

a = 17 - 16

a = 1

Therefore , the first term and common difference of AP are 1 and 8

Now , the 15th term of given AP will be

$\sf \mapsto a_{15} = 1+ 14 \times 8 \\ \\ \sf \mapsto a_{15} =113$

$\sf \therefore \underline{The \: 15th \: term \: of \: given \: AP \: is \: 113}$