1)The sum of the first 6 terms of an A.P is 42.The ratio of its10th to its 30"th terms is1:3.Find the first term and 13th term of the A.P

2)In an A.P the first term is 8. nth term is 33 and sum to first n terms is 123.Find n and d

3)In an AP prove that am+n+am-n=2am where an denotes nth term of the A.P

4)If the sum of first m terms of an AP is n and sum of first n terms is m,then show that the sum of first(m+n)terms is-(m+n)

5)Which term of the sequence 20, 77/4,37/2,71/3.... is the first negative term..

Refer the following links :
(1). https://www.meritnation.com/ask-answer/question/the-sum-of-first-6-terms-of-an-a-p-is-42-the-ratio-of-its/arithmetic-progressions/3955845

(2). We have, first term, a = 8
nth term , a= 33
Sn123
now,  a= 33 
     a + (n - 1)d = 33
      8 + (n - 1)d  = 33
  (n - 1) d = 25           ...............(1)
 Sn123 
 n22a + (n - 1)d = 123
n216 + (n - 1)d = 123n216 + 25 = 123                                     [using (1)]n = 123 × 241= 6
from (1), we have
(n - 1)d = 25
        5d = 25  so, d = 5

(3) . https://www.meritnation.com/ask-answer/question/1-in-an-ap-prove-that-am-n-am-n-2am-where-an-is-th/arithmetic-progressions/3618724

(4)https://www.meritnation.com/ask-answer/question/if-the-sum-of-first-m-terms-of-an-ap-is-n-and-the-sum-of-fi/arithmetic-progressions/1687556

(5). Refer the link for similar query :
 https://www.meritnation.com/ask-answer/question/which-term-of-the-sequence-17-161-5-152-5-143-5/arithmetic-progressions/3446548
 

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1)Given,a+ 9d:a+ 29d =1 : 3

= a+9d/a+29d = 1/3

on cross- multiplication,,

3a + 27d = a +29d

3a -a = 29d - 27d

2a = 2d

a =d.................... (i)

s6= 6/2 ( a + 5d) ,

42 = 3 (a + 5a) [ from (i)]

14 = 6a

14/6 = a = d

a13= a + 12 d

a13= 14/6 + 12 * 14/6

a13= 14/6 + 28

a13= 168+14/6 = 182/6 = 91/3

2) a =8

an=33

Sn= 123

Sn = n/2(a+an)

123 = n/2 x 41

123 x 2 /41 = n

6 = n

then, an = a+(n-1)d

33 = 8 + 5d

25 =5d

d = 5

therefore , n = 6 and d =5

3) Let the first term and the common difference of the A.P. be a and d respectively. Hence,LHS= am+n+am-n=2am= {a + (m + n 1)d} + {a + (m n 1)d}= 2a + (m + n 1 + m n 1)d= 2a + 2(m 1)d= 2(a + (m 1)d)=2am= RHS

4)A.T.Pn = m/2{2a+(m-1)d }

m = n/2{2a + (n-1)d}

n-m = m/2{2a+(m-1)d } - n/2{2a + (n-1)d}

n-m = am +m2d/2 - md/2 -an - n2d/2 + nd/2

n-m = a(m-n) + d/2 (m2- m -n2+n)

n-m= a(m-n) + d/2 [ (m+n) (m-n) -(m-n)]

n-m = a(m-n) + d/2[(m-n)(m+n-1)]

n-m = (m-n) [a + d/2(m+n-1)]

-(m-n) = (m-n) [a + d/2(m+n-1)]

-1 = 2a+d(m+n-1) upon 2

-2 = 2a+d(m+n-1)

since S(m+n)=m+n/2 * [2a+d(m+n-1)]

S(m+n) =m+n/2 * -2

S(m+n) = -(m+n)

Hope it helps...thumbs up plzz

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