Question

The sum of the first five terms of an AP is 25 and the sum of the terms from the sith to the fifteenth terms inclusive is 200. find the first term and the common difference

Answer 1

a = 1 and d = 2

Step-by-step explanation:

★Given -

• Sum of the first five terms of an AP is 25.
• Sum of the terms from the sixth to the fifteenth terms inclusive is 200.

★To find -

• First term(a).
• Common difference (d).

★Solution -

Formula for sum of 'n' terms of an A.P is,

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

Sum of first five terms is,

$\longmapsto \rm{ \bf{S {\tiny{5}} = \dfrac{5}{2} \{2a + (5 - 1)d \}} }$

$\longmapsto \rm{ \bf{25 = \dfrac{5}{2} \{2a + (4)d \}} }$

${\longmapsto \rm{ \bf{ \dfrac{{ \cancel{25}} \: \: ^{5} \times 2}{ \cancel5} = \{2a + 4d \}} }}$

${\longmapsto \rm{ \bf 5 \times 2= 2a + 4d }}$

${\longmapsto \rm{ \bf \underline \pink{10= 2a + 4d }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf (1)}$

Now, next 10 terms are from 6th term to 15th term.

${\longmapsto \rm{ \bf{200 = a {\tiny{6}} + a {\tiny{7}} + a {\tiny{8}} + a {\tiny{9}} + a {\tiny{10}}+a {\tiny{11}}+ a {\tiny{12}}+a{\tiny{13}}+ a {\tiny{14}}+a{\tiny{15}}}}}$

${\longmapsto \rm{ \bf{200 = a + 5d + a + 6d + a + 7d + a + 8d + a + 9d + a + 10d + a + 11d + a + 12d + a + 13d + a + 14d} }}$

${\longmapsto \rm{ \bf{200 = 10a + 95d} }}$

${\longmapsto \rm{ \bf{200 = 5(2a + 19d)} }}$

${\longmapsto \rm{ \bf{ \dfrac{ {{\cancel{200} \: \: ^{40} }}}{ \cancel5} = \{2a + 19d \}} }}$

${\longmapsto \rm{ \bf \underline \purple{ 40= 2a + 19d } \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: (2)}}$

(2) - (1),

${ \longmapsto \rm{ \bf{ 40 - 10=2a + 19d - (2a + 4d)} }}$

${ \longmapsto \rm{ \bf{ 30=2a + 19d - 2a - 4d} }}$

$\longmapsto \rm{ \bf{ 30= 15d} }$

$\longmapsto \rm{ \bf{ \dfrac{ \cancel{30} \: \: ^{2} }{ \cancel{15}} = d} }$

${ \longmapsto \rm{ \bf \orange{ \boxed{2 = \rm{ d}}} \: \: \: \: \: \: \: \: \: \: \: \{putting \: in \: (1) \}} }$

${\longmapsto \rm{ \bf 10 = 2a + 4(2)}}$

${\longmapsto \rm{ \bf 10 = 2a + 8}}$

${\longmapsto \rm{ \bf 10 - 8= 2a }}$

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

${\longmapsto \rm{ \bf \dfrac{ \cancel2 \: ^{1} }{ \cancel2 \: ^{1} } = a }}$

${\longmapsto {\rm{ \bf \blue{ \boxed{1 = \rm a}}}}}$

Therefore, a = 1 and d = 2.

Answer 2

Answer:

Step-by-step explanation: