if the 6th term of AP is -10 and its 10th term is -26 then find the 15th term of the AP About the author Maria

Answer: Given: the 6th term of AP is -10 10th term is -26 To find: the 15th term of the AP Solution: Let a be the first term and d be the common difference of the given AP. Then, [tex] \sf \: T_6=-10 \\ \\ \sf \implies \: a – 5d = – 10 \: \: – – – – – (i)[/tex] [tex] \sf \: T_10=-26 \\ \\ \sf \implies \: a + 9d = – 26 \: \: \: \: \: – – – – (ii)[/tex] On subtracting (I) from (ii) [tex] \sf \: 4d=(-26+10) \\ \\ \sf \implies4d=-16 \\ \\ \sf \implies d = \cancel \frac{ – 16}{4} \\ \\ \sf \implies \: d=-4[/tex] Putting d = –4 in (I) [tex] \sf \: a+5d=-10 \\ \\ \sf \implies \: a+5×(-4)=-10 \\ \\ \sf \implies \: a – 20 = – 10 \\ \\ \sf \implies \: a = – 10 + 20 \\ \\ \sf \implies \: a = 10[/tex] Now the 15th term of this AP [tex] \sf \: T_{15}=a+(15-1)d \\ \\ \sf \: = a + 14d \\ \\ \sf = 10 + 14 \times ( – 4) \\ \\ \sf = 10 – 56 \\ \\ \sf = – 46[/tex] Hence, the 15th term of this AP is –46. Reply

Answer:Given:To find:Solution:Let a be the first term and d be the common difference of the given AP. Then,

[tex] \sf \: T_6=-10 \\ \\ \sf \implies \: a – 5d = – 10 \: \: – – – – – (i)[/tex]

[tex] \sf \: T_10=-26 \\ \\ \sf \implies \: a + 9d = – 26 \: \: \: \: \: – – – – (ii)[/tex]

Onsubtracting(I)from(ii)[tex] \sf \: 4d=(-26+10) \\ \\ \sf \implies4d=-16 \\ \\ \sf \implies d = \cancel \frac{ – 16}{4} \\ \\ \sf \implies \: d=-4[/tex]Puttingd=–4in(I)[tex] \sf \: a+5d=-10 \\ \\ \sf \implies \: a+5×(-4)=-10 \\ \\ \sf \implies \: a – 20 = – 10 \\ \\ \sf \implies \: a = – 10 + 20 \\ \\ \sf \implies \: a = 10[/tex]Nowthe15thtermofthisAP[tex] \sf \: T_{15}=a+(15-1)d \\ \\ \sf \: = a + 14d \\ \\ \sf = 10 + 14 \times ( – 4) \\ \\ \sf = 10 – 56 \\ \\ \sf = – 46[/tex]Hence,the15thtermofthisAPis–46.