Q. 84.3( 12 Votes )

In an AP, if S<su

Answer :

Let the a be first term and d be common difference of an AP


Then sum of first n terms of an AP:



Given


S5 + S7 = 167



5[2a + 4d] + 7[2a + 6d] = 167(2)


10a + 20d + 14a + 42d = 334


24a + 62d = 334


12a + 31d = 167


12a = 167 - 31d [eqn 1]


Also


S10 = 235



5[2a + 9d] = 235


2a + 9d = 47


12a + 54d = 282


167 - 31d + 54d = 282 [using eqn 1]


23d = 115


d = 5


using this value in eqn 1


12a = 167 - 31(5)


12a = 167 - 155 = 12


a = 1


So, the AP is


a, a + d, a + 2d, a + 3d, … ,


1, 6, 11, 16


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