Question

V. Answer the following
34. The sum of 6 terms which form an AP is 345 and the difference between
the first and last term is 55. Find the terms.​

Answer 1

$\bold{\underline{\underline{\huge{\sf{AnsWer:}}}}}$

Terms in AP are :

• 30,41,52,63,74,85

$\bold{\underline{\underline{\large{\sf{StEp\:by\:stEp\:explanation:}}}}}$

GiVeN :

• Sum of 6 terms of an AP is 345
• Difference between the first term and last term is 55

To FiNd :

• The terms

SoLuTioN :

Let the six consecutive terms be :

1. a
2. a + d
3. a + 2d
4. a + 3d
5. a + 4d
6. a + 5d

$\sf{\underline{\underline{As\:per\:the\:first\:condition:}}}$

• Sum of terms = 345

$\hookrightarrow$ $\sf{a\:+\:a\:+\:d\:+\:a\:+\:2d\:+\:a\:+\:3d\:+\:a\:+\:4d\:+\:a\:+\:5d=\:345}$

$\hookrightarrow$ $\sf{6a\:+\:15d\:=\:345}$ ---> (1)

$\sf{\underline{\underline{As\:per\:the\:second\:condition:}}}$

• Difference between first and last term is 55

• First term : a
• Last term : a + 5d

$\hookrightarrow$ $\sf{a\:-\:a\:+\:5d\:=\:55}$

$\hookrightarrow$ $\sf{5d\:=\:55}$

$\hookrightarrow$ $\sf{d\:=\:{\dfrac{55}{5}}}$

$\hookrightarrow$ $\sf{d\:=\:11}$

Substitute d = 11 in equation 1,

$\hookrightarrow$ $\sf{6a\:+\:15d\:=\:345}$

$\hookrightarrow$ $\sf{6a\:+\:15\:\times\:11\:=\:345}$

$\hookrightarrow$ $\sf{6a\:+\:165\:=\:345}$

$\hookrightarrow$ $\sf{6a\:=\:345\:-\:165}$

$\hookrightarrow$ $\sf{6a\:=\:180}$

$\hookrightarrow$ $\sf{a\:=\:{\dfrac{180}{6}}}$

$\hookrightarrow$ $\sf{a\:=\:30}$

Terms in AP are :

1. a = 30
2. a + d = 30 + 11 = 41
3. a + 2d = 30 + 22 = 52
4. a + 3d = 30 + 33 = 63
5. a + 4d = 30 + 44 = 74
6. a + 5d = 30 + 55 = 85

Answer 2

$\bold{\underline{\underline{\huge{\sf{ANSWER\::}}}}}$

Given:

The sum of 6 terms which form an A.P. is 345 and the difference between the first and last term is 55.

To find:

The terms.

Explanation:

General term of Arithmetic Progression:

let a be the first term and d be the common difference of an A.P.

Then it is of the form, a,(a+d),(a+2d),(a+3d)......nth term= a+(n-1)d

According to the question:

There are only 6 terms;

• First term of an A.P.= a
• Second term of an A.P.= a+d
• Third term of an A.P.= a+2d
• Fourth term of an A.P.= a+3d
• Fifth term of an A.P.= a+4d
• Six term of an A.P.= a+5d

Sum of six term of an A.P. is 345

∴ a+a+d+a+2d+a+3d+a+4d+a+5d= 345

→ 6a +15d= 345....................(1)

Therefore,

Difference between the first term and last term is 55.

We have,

• First term= a
• Last term= a+5d

So,

Common difference,(d)= (a+5d) -(a)= 55

Common difference,(d)= $\cancel{a}$+5d $\cancel{-a}$=55

Common difference,(d)= 5d= 55

Common difference,(d)= d= 55/5

Common difference,(d)= d= 11.

Now,

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

⇒ 6a +15(11) = 345

⇒ 6a + 165 =345

⇒ 6a = 345 -165

⇒ 6a = 180

⇒ a= 180/6

⇒ a= 30

Thus,

→The first term of an A.P. is 30.

→The second term of an A.P. is 30 +11= 41.

→The third term of an A.P. is 30+2(11)=52.

→The fourth term of an A.P. is 30+3(11)=63.

→The fifth term of an A.P. is 30+4(11)= 74.

→The six term of an A.P. is 30+5(11)= 85.

Note:

• The nth term of A.P. is called general term.