Question

the 5th term of an A.P. is 8. The 8th term exceeds three times the 2nd term by 2 . find the first term, the common difference and the sum of first 15 term. For 95 point

Answer 1

a₅ = 8

a + 4d = 8 => Equation 1

a₈ = 3 ( a₂ ) + 2

=> a + 7d = 3 ( a + d ) + 2

=> a + 7d = 3a + 3d + 2

=> a - 3a + 7d - 3d = 2

=> - 2a + 4d = 2

Dividing by common factor 2 throughout the above equation, we get,

=> - a + 2d = 1 => Equation 2

Solving Equation 1 and Equation 2 we get,

a + 4d = 8

-a + 2d = 1

________

6d = 9 ( 'a' and '-a' gets cancelled. Hence we get only 6d = 9 by adding )

________

=> d = 9 / 6 = 3 / 2 = 1.5

a + 4d = 8

a + 1.5 ( 4 ) = 8

a + 6 = 8

a = 8 - 6

a = 2.

Hence a = 2 and d = 1.5.

Sum of first fifteen terms ( S₁₅ ) = 15 / 2 [ 2 ( 2 ) + ( 15 - 1 ) 1.5 ]

=> S₁₅ = 15 / 2 [ 4 + 14 * 1.5 ]

=> S₁₅ = 15 / 2 [ 4 + 21 ]

=> S₁₅ = 15 / 2 * 25

=> S₁₅ = 7.5 * 25 = 187.5

Hence the sum of the first fifteen terms is 187.5. The first term ( a ) is 2 and the common difference ( d ) is 1.5

Hope it helped !

Answer 2

Let 2nd term be x.

$T_5 = 8 \\ \\ T_2 = x \\ \\ T_8 = 3x + 2 \\ \\ \\ T_5 - T_2 = 8 - x \\ \\ = (5 - 2)d = 8 - x \\ \\ = 3d = 8 - x \\ \\ \\ T_8 - T_5 = 3x + 2 - 8 \\ \\ = (8 - 5)d = 3x - 6 \\ \\ = 3d = 3x - 6$

$8 - x = 3x - 6 \\ \\ 3x + x = 8 + 6 \\ \\ = 4x = 14 \\ \\ x = \frac{14}{4} = 3\frac{1}{2} = T_2 \\ \\ \\ 8 - x = 8 - 3\frac{1}{2} = 4\frac{1}{2} = 3d \\ \\ d = 4\frac{1}{2} \div 3 = \frac{9}{2} \times \frac{1}{3} = \frac{9}{6} = 1\frac{1}{2}$

We found the common difference.

$T_5 - T_1 = (5 - 1)d \\ \\ = 8 - T_1 = 4 \times 1\frac{1}{2} \\ \\ = 8 - T_1 = 6 \\ \\ \\ T_1 = 8 - 6 = 2$

We found the first term.

$T_{15} = T_1 + 14d \\ \\ = 2 + 14 \times 1\frac{1}{2} \\ \\ = 2 + 21 = 23 \\ \\ \\ S_{15} = \frac{15}{2} [T_1 + T_{15}] \\ \\ = \frac{15}{2} [2 + 23] \\ \\ = \frac{15 \times 25}{2} = \frac{375}{2} = 187.5$

We found the sum of first 15 terms.

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