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## Question

**Q1. What is the sum of the ﬁrst 7 terms of the series 1/3, 1/2, 3/4, …?**

- 2059/164
- 2050/192
- 2059/164
- None of these

**Q2. The sum of 15 terms of an AP is 600, and the common difference is 5. Find the ﬁrst term**

- 4
- 5
- 3
- None of these

**Sum of an AP [latex] = \frac{n}{2}\left ( 2a+\left( n-1 \right )d \right ) [/latex]**

**= 600= 15/2 (2a+ 14*5)**

**= By solving, we get a=5.**

**Q3. A man arranges to pay off a debt of `3600 by 40 annual installments which form an AP. When 30 of the installments were repaid, he died leaving a third of the debt unpaid. What is the value of the first installment**

**50****51****52****53**

**let, **

**First Installment = a, and common difference = d**

**= 3600=40/2 (2a+39d)**

**2a+39d= 180…..I**

**And, **

**30 installments were repaid, he died after living third debt unpaid,**

**= 3600/3 = 10/2 (2(a+30d)+9d)**

**= 1200 = 5 (2a+69d)**

**= 2a+69d = 240…..II**

**From Eq. I and II**

**a = 51, and d = 2**

**Q4. The first term in a sequence of integers is 2 and the second term is 10. All subsequent terms are the arithmetic mean of all of the preceding terms. What is the 39th term?**

**5****6****600****300**

**The first term and second term average out to 6. So the third term is 6. Now add 6 to the preceding two terms and divide by 3 to get the average of the first three terms, which is the value of the 4th term. This, too, is 6 (18/3)—all terms after the 2nd are 6, including the 39th. Thus, the answer is 6.**

**Q5. ****The sequence a _{1}+a_{2}+...+a_{n} begins with the numbers 3,11,18,... and has the n^{th} term defined as a_{1}+2n+n^{2}, for n≥2.**

**What is the value of the 20 ^{th} term of the sequence?**

**155****460****220****443**

**The first term of the sequence is a _{1}, so here a_{1}=3, and we’re interested in finding the 20^{th} term, so we’ll use n = 20.**

**Plugging these values into the given expression for the nth term gives us our answer**

**a1 + 2n ^{2}+n^{2}**

**a1=3 and n=20**

**3 + 2(20) + 20 ^{2}=443 **