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Q1. Find three terms in AP such that their sum is 3 and product is -8.

Solution

Let a - d, a and a + d be three terms in AP. According to the question, a - d + a + a + d = 3 3a = 3 or a = 1 (a - d) (a) (a + d) = -8 a(a² - d²) = -8 Putting the value of a = 1, we get, 1 - d² = -8 d² = 9 or d = ±3 Thus, the required three terms are -2, 1, 4 or 4,1,-2.

Q2. Find the sum of the first 50 odd natural numbers.

Solution

50 odd natural numbers are 1,3,5,...a_n Sn = n/2[2a+(n-1)d]= n/2(2x1 + 49x2) = 50/2 (100) = 2500

Q3. The 5^th term of an AP is 34 and the 15^th term is 9. Find the common difference and the r^th term of the AP.

Solution

Let 'a' and 'd' respectively be the first term and the common difference of the AP. According to the given information, t₅=a + 4d = 34 and t₁₅ = a + 14d = 9 Solving the two equations, we get a = 44; d = -2.5 Therefore, r^th term = a + (r-1)d = 44 -2.5(r-1) = 46.5 - 2.5 r

Q4. If the 3^rd and 6^th term of an AP are 7 and 13 respectively, find the 10^th term.

Solution

Here, t₃ = a + 2d = 7 .....(i) and t₆ = a + 5d = 13 .........(ii) Subtracting (i) from (ii) t₆ - t₃ = 5d - 2d = 6 3d = 6 d = 2 Substituting d = 2 in (i) a + 2× 2 = 7 a + 4 = 7 a = 3 Therefore, 10^th term = a + 9d = 3 + 9 x 2 = 21

Q5. Find how many two-digit numbers are divisible by 6.

Solution

The two-digit numbers that are divisible by 6 are 12, 18, 24 … …... 96 These numbers form an A.P. with first term 12 and common difference 6.  a = 12 and d = 6 Let the number of terms in this AP be n. Last term = an = 96 a + (n - 1)d = 96 12 + (n - 1) (6) = 96 2 + n - 1 = 16 n + 1 = 16 n = 16 - 1 = 15 Thus, 15 two-digit numbers are divisible by 6.

Q6. Vikram invested a certain amount in shares. The stock market, in that year, reached new heights throughout. The stock market grew at a rate such that his investment grew by a fixed amount every month. As a result, after four months, his investment became Rs. 10,000 and after nine months, it became Rs. 17,500. What amount will Vikram have after the twelfth month?

Solution

Let the initial amount invested by Vikram be Rs.a. Let his investment increase by Rs.d each month. Initial investment = a Thus, investment after the 1^st month = a + d Investment after the 2^nd month = a + 2d. Investment after the n^th month = a + nd   Investment after the 4^th month = a + 4d = 10,000  … (1) Investment after the 9^th month = a + 9d = 17,500  … (2)   Subtracting (1) from (2): 9d – 4d = 17,500 – 10,000 5d = 7,500 d = 1500   Substituting the value of d in (1): a + 4(1500) = 10,000 a = 10,000 – 6000 a = 4000   Thus, he invested Rs.4,000 in the stock market and his investment grew by Rs.1500 each month. Thus, after the 12^th month, his investment became a + 12d = Rs. 4,000 + 12 x (1500) = Rs. 4,000 + 18,000 = Rs. 22,000

Q7. The mth term of an AP is n and the nth term is m. Find the rth term of the AP.

Solution

Let a and d respectively be the first term and the common difference of the AP. According to the given conditions, a+(m-1)d =n ...(i) a+(n-1)d =m ...(ii) On solving (i) and (ii), we get, d = -1 ; a= m+n-1 Therefore, r^th term = a+(r-1)d = (m+n-1) - (r-1) = m+n-r

Q8. Find the number of terms of the series: -5 + (-8) + (-11) + ...... + (-230)

Solution

-5 + (-8) + (-11) + ...... + (-230) The series -5, -8, -11, … -230 is an arithmetic progression with First term, a = -5 Common difference, d = -3 Let a_n = -230 -230 = a + (n - 1)d -230 = -5 + (n - 1)(-3) -230 = -5 - 3n + 3 -228 = -3n n = 76 Thus, there are 76 terms in the given sequence.

Q9. An A.P. consists of 60 terms. If the first and the last term be 7 and 125 respectively, find the 32^nd term.

Solution

Here, n = 60, a = 7 and ^t60 = 125 7+59d = 125 d=2 Therefore, 32^nd term (t₃₂) = a + 31d = 7 + 31 x 2 = 69

Q10. In a flower bed, there are 31 rose plants in the first row, 28 in the second, 25 in the third, and so on. There are 7 rose plants in the last row. How many rows are there in the flower bed?

Solution

The number of rose plants in the 1st, 2nd, 3rd … rows are 31, 28, 25 … 7                          

These numbers are in an AP. Let the number of rows in the flower bed be n. Then a = 31, d = 28 – 31 = – 3, a_n = 7                                                 

a_n = a + (n – 1) d                                                                       

Thus, 7 = 31 + (n – 1)(– 3) i.e., – 24 = (n – 1)(– 3) i.e.  8 = n - 1 i.e., n = 9 So, there are 9 rows in the flower bed.