In an AP:

(i) givenfind and


(ii) givenfind and


(iii) given find and


(iv) givenfind and


(v) givenfind a and


(vi) givenfind and


(vii) given find and


(viii) given find and


(ix) given find


(x) givenand there are total 9 terms. Find

(i) Number of terms can be calculated as follows:


an = a + (n – 1)d


Or, 50 = 5 + (n – 1)3


Or, (n – 1)3 = 50 – 5 = 45


Or, n – 1 = 15


Or, n = 16


Sum of n terms can be given as follows:



S16


= (10 + 45)


=


=440


(ii) Common difference can be calculated as follows:


an = a + (n – 1)d


Or, 35 = 7 + 12d


Or, 12d = 35 – 7 = 28


Or, d = 7/3


Sum of n terms can be given as follows:



S13


= (14 + 28)


=


=273


(iii) First term can be calculated as follows:


an = a + (n – 1)d


Or, 37 = a + 11 x 3


Or, a = 37 – 33 = 4


Sum of n terms can be given as follows:



S12


= 6(8 + 33)


= 6 * 41


= 246


(iv) Sum of n terms can be given as follows:



S10


125 = 5(2a + 9d)


25 = 2a * 9d (i)


According to the question; the 3rd term is 15, which means;


a + 2d = 15 (ii)


Now,


Subtracting equation (ii) from equation (i), we get;


2a + 9d – a – 2d = 25 – 15


Or, a + 7d = 10 (iii)


Subtracting (ii) from equation (iii), we get;


a + 7d – a – 2d = 10 – 15


Or, 5d = - 5


Or, d = - 1


Now,


Substituting the value of d in equation (2), we get;


a + 2(- 1) = 15


Or, a – 2 = 15


Or, a = 17


10th term can be calculated as follows;


a10 = a + 9d


= 17 – 9 = 8


Thus, d = - 1 and 10th term = 8


(v) Sum of n terms can be given as follows:



75


75= (2a + 40)


= 2a * 40



Now,


9th term can be calculated as follows:


a9 = a + 8d


= -


=


(vi) Sum of n terms can be given as follows:



90


90= (4 + 8N -8)


90 = N(2 +4N -4)


4N2 -2N -90 =0


2N2 -N -45 =0


(2N +9)(N -5)


Hence, n = - 9/2 and n = 5


Rejecting the negative value,


We have n = 5


Now, 5thterm will be:


a5 = a + 4d


= 2 + 4 x 8


= 2 + 32 = 34


(vii) Sum of n terms can be given as follows:



210


420 = (8 + 62)


420 = n *70


n = 6


Now, for calculating d:


a6 = a + 5d


Or, 62 = 8 + 5d


Or, 5d = 62 – 8 = 54


Or, d = 54/5


(viii) Sum of n terms can be given as follows:



-14


-28 = (a + 4)


(i)


We know;


an = a + (n – 1)d


Or, 4 = a + (n – 1)2


Or, 4 = a + 2n – 2


Or, a + 2n = 6


Or, 2n = 6 – a


Or, n = (6 – a)/2 (ii)


Using (i) and (ii):



-56 = (6 –a)(a +4)


24 + 2a – a2 = -56


a2 -2a -80= 0


(a + 8)(a – 10)= 0


Therefore, a = - 8 and a = 10


As a is smaller than 10 and d has positive value, hence we’ll take a = - 8


Now, we can find the number of terms as follows:


an = a + (n – 1)d


4 = - 8 + (n – 1)2


(n – 1)2 = 4 + 8 = 12


n – 1 = 6Or, n = 7


Hence, n = 7 and a = - 8


(ix) Sum of n terms can be given as follows:



192


192 = (6 + 7d)


7d = 42


d = 6


(x) Sum of n terms can be given as follows:



144 (a + an)


288 = (a + 28)


9a + 252 = 288


9a = 288 -252


9a = 36


a = 4


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