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**Hots questions****Q1. Find the â€˜6thâ€™ term of the A.P. : ****Sol.** Here,

âˆ´ d = a_{2} â€“ a_{1}

Now, a_{n} = a + (n â€“ 1)d

Thus, the n^{th} term is

Again, we have

â‡’

i.e., the 6^{th} term is **Q2. If the ratio of the sum of first n terms of two A.P.'s is (7n + 1) : (4n + 27), find the ratio of their m ^{th} terms.**

âˆ´

âˆ´

Now, replacing n by (2m-1) for getting ratio of mth terms of given APs,

Thus, the required ratio of mth term of given AP's is (14m â€“ 6) : (8m + 23)

Let D be the common difference of the given A.P.,

Then :

b = a + D, c = a + 2D, d= a + 3D and e = a + 4D

âˆ´ a â€“ 4b + 6c â€“ 4d + e

= a â€“ 4(a + D ) + 6 (a + 2D) â€“ 4 (a + 3D) + (a + 4D)

= a â€“ 4a + 6a â€“ 4a + a â€“ 4D + 12D â€“ 12D + 4D

= 8a â€“ 8a + 16D â€“ 16D = 0

Thus, a â€“ 4b + 6c â€“ 4d + e = 0

(a + b)

We know that :

A.M. between â€˜aâ€™ and â€˜bâ€™ = a+b/2

It is given that,

is the A.M. between â€˜aâ€™ and â€˜bâ€™

âˆ´

By cross multiplication, we get :

â‡’2a

â‡’ 2a

â‡’ a

â‡’ a

â‡’ a

â‡’

â‡’

â‡’ n = 0**Q5. Solve the equation :****1 + 4 + 7 + 10 + ... + x = 287 ****Sol.**

Since,

âˆ´ a = 1,d = 3 and a_{n} = x

âˆ´ a_{n} = a + (n â€“ 1)d

â‡’ x = 1 + (n â€“ 1) 3 or x = 3n â€“ 2

Also, S_{n} = n/2 (a+l)

â‡’

â‡’ 2(287) = n[1 + (3n â€“ 2)]

â‡’ 574 = n[3n â€“ 1]

â‡’ 3n^{2} â€“ n â€“ 574 = 0

Solving the above quadratic equation, we get

or

But, negative n is not desirable.

âˆ´ n = 14

x = 3n â€“ 2

Now, x = 3(14) â€“ 2 = 42 â€“ 2 = 40

Thus, x = 40**Q6. Find three numbers in A.P. whose sum is 21 and their product is 231.****Sol. **Let the three numbers in A.P. are:

a â€“ d, a, a + d

âˆ´ (a â€“ d) + a + (a + d) = 21

â‡’ a â€“ d + a + a + d = 21

or 3a = 21 â‡’ a = 7

Also, (a â€“ d) Ã— a Ã— (a + d) = 231

âˆ´ (7 â€“ d) Ã— 7 Ã— (7 + d) = 231

â‡’ (7 â€“ d) (7 + d) Ã— 7 = 231

â‡’ 7^{2} â€“ d^{2} = 231 /7 = 33

â‡’ 49 â€“ d^{2} = 33

or d^{2} = 49 â€“ 33 = 16

â‡’ d = Â± 4

Now, when d = 4, then three numbers in AP are : (7 â€“ 4), 7, (7 + 4) i.e. 3, 7, 11.

When d = â€“4, then three numbers in AP are : [7 â€“ (â€“4)], 7, [7 + (â€“4)]

or 11, 7, 3**Q7. The ninth term of an A.P. is equal to seven times the second term and twelfth term exceeds five times the third term by 2. Find the first term and common difference.****Sol.** Let 'a' be the first term and 'd' be the common difference

Since, an = a + (nâ€“1)d

âˆ´ a_{9} = a + 8d and a_{12} = a + 11d

Also a_{2} = a + d and a_{3} = a + 2d

since a_{9} = 7a_{2} and a_{12} = 5a_{3} + 2

or a_{9} = 7 (a + d) â‡’ a_{9} = 7a + 7d ... (1)

a_{12} = 5 (a + 2d) + 2

â‡’ a_{12} = 5a + 10d + 2 ... (2)

Now, a + 8d = 7a + 7d [From (1)]

or â€“ 6a + d = 0 ... (3)

Also a + 11d = 5a + 10d + 2 [From (2)]

or â€“4a + d = 2 ... (4)

Subtracting (4) from (3), we have

â€“2a = â€“2 â‡’ a = 1

Now, from (3), â€“6 + d = 0 â‡’ d = 6

Thus, a = 1 and d = 6

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