Question

The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th term, find the AP.

Answer 1

HELLO DEAR,

let first term be a.

common difference be d.

according to question,

CASE- 1

$\bold{a_2 + a_7 = 30}$

$\bold{\implies \{a + (2 - 1)d\} + \{a + (7 - 1)d\} = 30}$

$\bold{\implies 2a + 7d = 30-----------( \;1 \;)}$

CASE-2

$\bold{a_{15} = 2a_8 - 1}$

$\bold{\implies \{a + (15 - 1)d\} = 2\{a + (8 - 1)d\} - 1}$

$\bold{\implies 2a - a + 14d - 14d - 1 = 0}$

$\bold{\implies a = 1}$ [put in ----( 1 )]

we get,

2( 1 ) + 7d = 30

7d = 30 - 2

7d = 28

d = 28/7

d = 4

thus, a = 1 , d = 4

so, the arithmetic series is; 1 , (1 + 4) , (1 + 2*4) , (1 + 3*4) , (1 + 4*4).....

1 , 5 , 9 , 13 , 17.......

I HOPE IT'S HELP YOU DEAR,

THANKS

Answer 2

$\huge \bold {\mathfrak {hey}}$

$\huge \bold {Solutions :-}$

➡️ let the first term be x

➡️ Difference be d

$\underline \bold {ATQ}$

In Case 1

a2 + a7 = 30

=> {a + ( 2 - 1 ) d } + { a + 7 - 1 ) d } = 30

➡️ 2a + 7d = 30 - - - - - - - ( 1 )

Now,

In case 2

a15 = 2a8 - 1

➡️ {a + ( 15 - 1 ) d = 2 { a + ( 8 - 1 ) d} - 1

=> 2a - a + 14d - 14d - 1 = 0

Therefore,

Value of a = 1

Put this value in equation ( 1 )

We get,

a(1) + 7d = 30

7d = 28
d = 28 / 7
d = 4

Therefore,

a =1 and d = 4

So the arithmetic series

1, (1+4), ( 1+2×4), (1 +3×4),(1+4×4)......

1,5,9,13,17 ✔️✔️✔️✔️



HOPE IT WILL HELP YOU



BE BRAINLY