Answer :

Given:

T11: T18 = 2: 3


We know that,


nth term = Tn = a + (n – 1)d


where,


a = first term


d = common difference



T11 = a + (11 – 1)d


T11 = a + 10d …(1)


And,


T18 = a + (18 – 1)d


T18 = a + 17d …(2)


From (1) and (2),



3(a + 10d) = 2(a + 17d)


3a + 30d = 2a + 34d


3a 2a = 34d 30d


a = 4d (3)


We know that,



Now,




From (3),







Hence, S5:S10 = 6:17


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