Question

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1. In an A.P 1,7, 13,19,…………..,415. Prove that the sum of the middle terms is equal to the sum of first and last terms
2. If the 8th term of an A.P is 31 and the 15th term is 16 more than the 11th term, find the A.P.
3. Which term of the A.P 5,15,25, ……….. will be 130 more than its 31st term.
4. If the 10th term of an A.P is 52 and its 17th term is 20 more than its 31st term. Find the A.P.
5. Is 184 a term of the A.P 3,7,11,……..
6. The sum of 5th and 9th terms of an A.P is 72 and the sum of 7th and 12th terms is 97. Find the A.P
7. If 5 times the fifth term of an A.P is equal to 8 times its eighth term, show that its 13th term is zero.
8. How many numbers of two digits are divisible by 7
9. Find the number of integers between 50 and 500 which are divisible by 7
10. If the mth term of an A.P be 1/n and nth term be 1/m , then show that its (mn)th term is 1.
11. If the pth term of an A.P is q and the qth term is p, prove that its nth term is (p+q-n).
12. If m times the mth term of an A.P is equal to n times its nth term, Show that the (m+n)th term of the A.P is zero
13. If pth , qth, rth terms of an A.P are a, b, c respectively,then show that a(q-r)+b(r-p)+c(p-q)=0
14. Which term of the sequence 20,19 1/4,181/2,173/4,…………… is the first negative term.
15. Find a, b and c if it is given that the numbers a,7,b,23,c are in A.P

Answer 1

Step-by-step explanation:

1)In a AP 1, 7, 13, 19, ....415, we have

a = 1, d = second term - first term = 7 -1 = 6 and last term = l =415.

So,

l = a+(n−1)d

⇒ 415 = 1+(n−1)6

⇒ 415 -1 = (n−1)6

⇒ 414 = (n−1)6

⇒ 414 = 6n − 6

⇒ 414 + 6 = 6n

⇒ 420 = 6n

⇒

⇒n=70

Clearly its middle terms are 35th and 36th,

Sum of middle terms =  = 205+211 =416       [1]

And sum of first and last term = a + l = 1 + 415 = 416                             [2]

From eq 1 and 2,

Sum of the middle terms is equal to the sum of first and last terms .

Hence proved.

2)The 8th term of the AP = a + 7d = 31 …(1)

The 11th term of the AP = a + 10d

The 15th term being 16 more than the 11th term]

The 15th term of the AP = a + 14d = 16 + a + 10d, or

14d - 10d = 16 or 4d = 16, or

d = 4.

From (1), a = 31–28 = 3.

The AP is 3, 7, 11, ….

Check: T11 = a+10d = 3+40 = 43.

T15 = a+14d = 3+56 = 59.

T15-T11 = 59–43 = 16. Correct.

3)Ap is 5,15,25.......

a=5 and d=15-5=10

31st term=a+30d

=5+30*10

=5+300

31st term =305

an=305+130

=435

a+(n-1)d=435

5+(n-1)10=435

(n-1)10=435-5

(n-1)=430/10

n-1=43

n=44

It means the 44th term of the given AP will be 130 more than its 31st term.

4)a+9d=52.........[1]

given that-(a+16d)-(a+12d)=20

which gives 4d=20

d=4

[keeping the value in eq, 1]

a+9×5=52

a=52-45

a=7

now we can make the Apple as 7,12,16.....

5)Let 184 be the n th term of the sequence.

a = 3, d = 7-3= 4 

therefore,   184 = a + (n-1) d

                         = 3 + (n-1) 4

184-3 = (n-1) 4

181 = (n-1) 4

n-1 = 181/4

n = 181/4 +1

n = 185/4

However, n should be a positive integer. Therefore, 184 is not a term of the given A.P.

6)Let of the AP

first term= a

common difference=d

ATQ

5[a+(5-1)d]=8[a+(8-1)d]

5(a+4d)=8(a+7d)

5a + 20d = 8a + 56d

3a = 36d

a= -12d

Now 13th term

a+(13-1)d

=-12d+12d

=0

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