Question

11. Which term of the AP:3, 15, 27, 39,... will be 132 more than its 54th term?
12. Two APs have the same common difference. The difference between their 100th terms is
100, what is the difference between their 1000th terms?
13. How many three-digit numbers are divisible by 7?​

Answer 1

Answer:

Step-by-step explanation:

Solutions :-

Answer 11.

Here, a = 3, d = 15 - 3 = 12 and n = 54

a(n) = a + (n - 1)d

⇒ a(54) = 3 + (54 – 1)12

⇒ a(54)  = 3 + 53 × 12

⇒ a(54)  = 3 + 636

⇒ a(54)  = 639

132 more than its 54th term = 132 + 639 = 771

Again, a(n) = a + (n - 1)d

⇒ a + (n - 1)d = 771

⇒ 3 + (n - 1)12 = 771

⇒ 3 + 12n - 12 = 771

⇒ 12n = 771 + 12 - 3

⇒ 12n = 780

⇒ n = 780/12

⇒ n = 65

Hence, the 65th term is 132 more than the 54th term.

Answer 12.

Let a, a' be the first term and d be the common difference of given A.P.

According to the Question,

⇒ a + 99d - a' - 99d = 100

⇒ a - a' = 100 ...... (i)

⇒ a + 999d - a' - 999d

= a - a'  (From (i))

= 100

Hence, the  difference between 1000th term is 100.

Answer 13.

A.P. is 105, 112, 119,......, 994.

Here, a = 105, d = 112 - 105 = 7

a(n) = 994

a(n) = a + (n - 1)d

⇒ 994 = 105 + (n - 1)7

⇒ 994 - 105 = 7(n - 1)

⇒ 889 = 7(n - 1)

⇒ n - 1 = 889/7

⇒ n - 1 = 127

⇒ n = 127 + 1

⇒ n = 128.

Hence, 128 three digits numbers are divisible by 7.

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Answer 2

Solution 11:

$\bf{\red{\underline{\bf{Given\::}}}}$

The A.P. is 3,15,2,39..... will be 132 more than 54th term.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

Which term.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

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.

So, we have;

• First term (a) = 3
• Common difference (d) = 12
• nth term = 54

A/q

$\longrightarrow\sf{a_{54}=3+(54-1)12}\\\\\longrightarrow\sf{a_{54}=3+53\times 12}\\\\\longrightarrow\sf{a_{54}=3+636}\\\\\longrightarrow\sf{\pink{a_{54}=639}}$

&

Term = 132 + 54th

Term = 132 + 639

Term = 771

$\longrightarrow\sf{a_{n}=3+(n-1)12}\\\\\longrightarrow\sf{771=3+12n-12}\\\\\longrightarrow\sf{771=12n-9}\\\\\longrightarrow\sf{12n=771+9}\\\\\longrightarrow\sf{12n=780}\\\\\longrightarrow\sf{n=\cancel{\dfrac{780}{12} }}\\\\\\\longrightarrow\sf{\pink{n=65}}$

Thus;

The nth term is 65 .

Solution 12:

$\bf{\red{\underline{\bf{Given\::}}}}$

Two A.P. have the same common difference. The difference between their 100th term is 100.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The difference between their 1000th terms.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We know that formula;

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

We have two type of A.P;

$\bullet\sf{A.P_{1}=First\:term\:(a)_1\:\:\:\:\:\:\&\:\:\:\:common\:difference\:(d)_1}}\\\bullet\sf{A.P_{2}=First\:term\:(a)_2\:\:\:\:\:\:\&\:\:\:\:common\:difference\:(d)_2}}$

common difference same;

$\sf{d_{1}=d_{2}=d}$

A/q

$\longrightarrow\sf{a_{1}100-a_{2}100=100}\\\\\longrightarrow\sf{\big[a_{1}+(100-1)d_{1}\big]-\big[a_{2}+(100-1)d_{2}\big]=100}\\\\\longrightarrow\sf{a_{1}+99d-\big[a_{2}+99d\big]=100}\\\\\longrightarrow\sf{a_{1}\cancel{+99d}-a_{2} \cancel{-99d}=100}\\\\\longrightarrow\sf{\pink{a_{1}-a_{2}=100............(1)}}$

Now;

$\longrightarrow\sf{a_{1}1000-a_{2}1000}\\\\\longrightarrow\sf{\big[a_{1}+(1000-1)d_{1}\big]-\big[a_{2}+(1000-1)d_{2}\big]}\\\\\longrightarrow\sf{a_{1}+999d_{1}-\big[a_{2}+999d_{2}\big]}\\\\\longrightarrow\sf{a_{1}\cancel{+999d}-a_{2} \cancel{-999d}}\\\\\longrightarrow\sf{\pink{a_{1}-a_{2} = 100}}\:\:\:\:[from (1)]}$

Thus;

The common difference their 1000th term is 100 .

Solution 13:

$\bf{\red{\underline{\bf{Given\::}}}}$

The number divisible by 7.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The three digit number.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

The three number divisible by 7,we get;

105,112,119,126..................994

$\underline{\underline{\bf{According\:to\:the\:question\::}}}}}$

We know that formula of an A.P.

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

• a is the first term = 105
• l is the last term = 994
• d is the common difference = 112 - 105 = 7

Therefore;

$\longrightarrow\sf{a_{n}=105+(n-1)7}\\\\\longrightarrow\sf{a_{n}=105+7n-7}\\\\\longrightarrow\sf{a_{n}=7n+98....................(1)}$

Then;

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

$\longrightarrow\sf{a_{n}=994}\\\\\longrightarrow\sf{7n+98=994}\\\\\longrightarrow\sf{7n=994-98}\\\\\longrightarrow\sf{7n=896}\\\\\longrightarrow\sf{n=\cancel{\dfrac{896}{7} }}\\\\\\\longrightarrow\sf{\pink{n=128}}$

Thus;

The three digit number (128) which are divisible by 7 .