Question

in an a.p 6th term is one more than twice the 3rd term .the sum of the 4th and 5th term is 5 time the 2nd term find the 10th term of the a.p

Answer 1

Answer:

29

Step-by-step explanation:

Let the first term be a and common difference be d.

According to question:

= > 6th term = 1 + twice of 3rd term

= > a + 5d = 1 + 2( a + 2d )

= > a + 5d = 1 + 2a + 4d

= > 5d - 4d - 1 = 2a - a

= > d - 1 = a ... (1)

Whereas,

= > 4th term + 5th term = 5 times of 2nd term

= > a + 3d + a + 4d = 5[ a + d ]

= > 2a + 7d = 5a + 5d

= > 7d - 5d = 5a - 2a

= > 2d = 3a

= > 2d = 3( d - 1 ) { from (1) }

= > 2d = 3d - 3

= > 3 = 3d - 2d

= > 3 = d

Hence, a = d - 1 = 3 - 1 = 2

Therefore,

= > 10th term

= > a + 9d

= > 2 + 9(3)

= > 2 + 27

= > 29

Answer 2

Solution :

Firstly, we know that formula of an A.P;

$\boxed{\bf{a_n=a+(n-1)d}}}$

• a is the first term.
• d is the common difference.
• n is the term of an A.P.

A/q

$\longrightarrow\sf{a_6=1+2(a_3)}\\\\\longrightarrow\sf{a+(6-1)d = 1+2(a+3-1)d}\\\\\longrightarrow\sf{a+5d= 1+2(a+2d)}\\\\\longrightarrow\sf{a+5d=1+2a+4d}\\\\\longrightarrow\sf{a-2a+5d-4d=1}\\\\\longrightarrow\sf{-a+d=1}\\\\\longrightarrow\sf{d=1+a}\\\\\longrightarrow\bf{a=d-1...................(1)}$

&

$\longrightarrow\sf{a_4 + a_5 = 5(a_2)}\\\\\longrightarrow\sf{a+(4-1)d + a+(5-1)d = 5[a+(2-1)d]}\\\\\longrightarrow\sf{a+3d + a + 4d = 5(a+d)}\\\\\longrightarrow\sf{2a + 7d = 5a + 5d}\\\\\longrightarrow\sf{5a - 2a + 5d-7d=0}\\\\\longrightarrow\sf{3a - 2d=0}\\\\\longrightarrow\sf{3(d-1) - 2d =0\:\:[from(1)]}\\\\\longrightarrow\sf{3d- 3 - 2d = 0}\\\\\longrightarrow\sf{-3+d=0}\\\\\longrightarrow\bf{d=3}$

∴ Putting the value of d in equation (1),we get;

$\longrightarrow\sf{a=3-1}\\\\\longrightarrow\bf{a=2}$

Now;

$\longrightarrow\sf{a_{10} = a+(10-1)d}\\\\\longrightarrow\sf{a_{10} = 2 + 9(3)}\\\\\longrightarrow\sf{a_{10} = 2+27}\\\\\longrightarrow\bf{a_{10}=29}$

Thus;

The 10th term of an A.P. will be 29 .