Question

sum of the first 9th term is 261 and next 6 terms is 444 find 1st term

Answer 1

5

Step-by-step explanation:

★Correct question

• Consider an arithmetic sequence whose sum of first 9 terms is 261 and sum of next 6 terms is 444.

★Given -

• Sum of first 9 terms is 261.
• Sum of next 6 terms is 444.

★To find -

• The 1st term.

★Solution -

Formula for sum of n terms in an A.P is

$\rm{ \bf \blue{S {\tiny{n}} = \dfrac{n}{2} \{2a + (n - 1)d \} }}$

where, a = first term of the A.P

and, d = common difference

On putting the values in the formula, we get,

$\longmapsto \rm{{261} = \dfrac{9}{2} \{2a + (n - 1)d \} }$

$\longmapsto \rm{{ \dfrac{ {\cancel{261} \: ^{29} }\times 2}{ \cancel9} } = 2a + (9 - 1)d }$

$\longmapsto \rm{{ 29 \times 2 } = 2a + (8)d }$

$\longmapsto \rm{ \bf \pink{58 = 4a + 8d} \: \: \: \: \: \: \: \:\:\: \:\: \: \: \:\: \: \: \: \: \: \: \: (1)}$

Now, the next 6 terms after 9th term is,

=> 10th, 11th, 12th, 13th, 14th, 15th.

Sum of next 6 terms is,

${ \longmapsto\rm{ a {\tiny10} + a {\tiny11} + a {\tiny12} + a {\tiny13} + a {\tiny14} + a {\tiny15} = 441 }}$

${ \longmapsto \rm{a + 9d + a + 10d + a + 11d + a + 12d + a + 13d + a + 14d = 441}}$

$\longmapsto \rm{6a + 69d = 444}$

$\longmapsto \rm{(2a + 23d)3 = 444}$

${ \longmapsto \rm{ \underline{ \purple{\bf2a + 23d = 148}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (\bf 2)}}$

(2) - (1),

${ \longmapsto \rm{148 - 58= 2a + 23d - (2a + 8d)}}$

${ \longmapsto \rm{90= 2a + 23d - 2a - 8d}}$

${ \longmapsto \rm{90= 15d}}$

${ \longmapsto \rm{\dfrac{\cancel{90}}{\cancel{15}}= d}}$

${ \longmapsto \rm{\bf \red{6= d}}\:\:\:\:\:\:\{putting \: in\: (1)\}}$

${ \longmapsto \rm{58= 2a + 8(6)}}$

${ \longmapsto \rm{58= 2a + 48}}$

${ \longmapsto \rm{58 - 48 = 2a }}$

${ \longmapsto \rm{10 = 2a }}$

${ \longmapsto \rm{\dfrac{\cancel{10}}{\cancel{2}}= d}}$

${ \longmapsto \rm{\bf \green{5 = a} }}$

Therefore, the 1st term is 5.